AVIATION MAINTENANCE TRAINING CENTRE
MT.147.02
M1 MATHEMATICS
EASA PART-66 CAT B1/B2 ISSUE: 1JAN2007
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Training Manual
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by Lufthansa Technical Training (LTT).
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MATHEMATICS
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1.
ARITHMETIC
1.1 General Just as studying a new language begins with learning basic words, the study of mathematics begins with arithmetic, its most basic branch. Arithmetic uses real and non−negative numbers, which are also known as counting numbers, and consist of only four operations, addition, subtraction, multiplication and division. While you have been using arithmetic since childhood, a review of its terms and operations will make learning the more difficult mathematical concepts much easier. Numbers are represented by symbols which are called digits. There are nine digits which are 1, 2, 3, 4, 5, 6, 7, 8, & 9. We also use the symbol 0 (ie zero) where no digits exists. Digits and zero may be combined together to represent any number.
1.2 Addition The process of finding the total of two or more numbers is called addition. This operation is indicated by the plus (+) symbol. When numbers are combined by addition, the resulting total is called the sum. When adding whole numbers whose total is more than nine, it is necessary to arrange the numbers in columns so that the last digit of each number is in the same column. The ones column contains the values zero through nine, the tens column contains multiples of ten, up to ninety, and the hundreds column consists of multiples of one hundred.
Example: L N O S E S O P R U P G N I N I A R T R O F
hundreds 2 + 4 7
tens 7 4 6 8
ones 8 3 2 3
To add the sum of the above, first add the ones column, 8 and 3 make 11 and 2 makes 13. Place the 3 in the ones column of the answer and carry the 1 forward to the tens column. Adding this we have 1 and 7 is 8 and 4 is 12 and 6 is 18. Place the 8 in the tens column of the answer and carry the 1 forward to the hundreds column which we now add. 1 and 2 is 3 and 4 is 7. Place the 7 in the hundreds column of the answer. We see that the answer (sum) to the addition is 783. The process is identical if any of the numbers includes a decimal as long as the decimal points are arranged in the same column. The number of digits after the decimal point as no significance.
Example: hundreds 2 +4 7
tens 7 4 6 8
. . . .
ones 8 3 0 3
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1.3 Subtraction The process of finding the difference between two numbers is known as subtraction and is indicated by the minus (−) sign. Subtraction is accomplished by taking the quantity of one number away from another number. The number which is being subtracted is known as the subtrahend (smaller number), and the number from which the quantity is taken is known as the minuend (larger number). To find the difference of two numbers, arrange them in the same manner used for addition. With the minuend on top and the subtrahend on the bottom, align the vertical columns so the last digits are in the same column. Beginning at the right, subtract the subtrahend from the minuend. Repeat this for each column. Example:
hundreds 4 −2 1
tens 4 6 8
ones 3 − minuend 2 − subtrahend 1
Place 262 under 443. 2 from 3 leaves 1. write 1 in the ones column of the answer. 6 from 4 is clearly impossible, so the 4 is increased in value to 14 by taking 1 from the hundreds column leaving 3. 14 from 6 leaves 8. Write 8 in the tens column. finally, 3 from 2 in the hundreds columns leaves 1. To check a subtraction problem, you can add the difference to the subtrahend to find the minuend. There are two methods by which subtraction can be performed. Consider
15 − 8 = 7 L N O S E S O P R U P G N I N I A R T R O F
1st method: take 8 from 15. We have 7 left. 2nd method: if to 7 we add 8 then we obtain 15. 7 is therefore the difference between 15 and 8. The process is identical if any of the numbers include a decimal as long as the decimal points are arranged in the same column.
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1.4 Multiplication Multiplication is a special form of repetitive addition. When a given number is added to itself a specified number of times, the process is called multiplication. The sum of 4 + 4 + 4 = 12 is expressed by multiplication as 4 x 3 = 12. The numbers 4 and 3 are called factors and the answer, 12, represents the product. The number multiplied (4) is called the multiplicand, and the multiplier represents the number of times the multiplicand is added to itself. Multiplication is typically indicated by an (x), ( ), or in certain equations, by the lack of any other operation sign. One important factor to remember when multiplying is that the order in which numbers are multiplied does not change the product. Example:
3 x4 12
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or
4 x3 12
Like addition and subtraction, when multiplying large numbers it is important they be aligned vertically. Regardless of the number of digits in the multiplicand or the multiplier, the multiplicand should be written on top, and the multiplier beneath it. When multiplying numbers greater than nine, multiply each digit in the multiplicand by each digit in the multiplier. Once all multiplicands are used as a multiplier, the products of each multiplication operation are added to arrive at a total product. Example:
532 − Multiplicand x 24 − Multiplier 10640 − First partial product 2128 − Second partial product 12,768 − Product
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1.5 Division Just as subtraction is the reverse of addition, division is the reverse of multiplication. Division is a means of finding out how many times a number is contained in another number. The number divided is called dividend, the number you are dividing by is the divisor, and the result is the quotient. With some division problems, the quotient may include a remainder. A remainder represents that portion of the dividend that cannot be divided by the divisor. Division is indicated by the use of the division sign () with the dividend to the left and the divisor to the right of the sign, or with the dividend inside the sign and the divisor to the left. Division also is indicated in fractional form. For example, in the fraction the 3 is the dividend and the 8 is the divisor. When division is carried out, the quotient is 0.375. The process of dividing large quantities is performed by breaking the problem down into a series of operations, each resulting in a single digit quotient. This is best illustrated by example. Example:
dividend divisor
416
8
= 52
or 52
8 ) 416 40 16 16 L N O S E S O P R U P G N I N I A R T R O F
To check a divsion problem for accuracy, multiply the quotient by the the divisor and add the remainder (if any). If the operation is carried out properly, the result equals the dividend.
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M1 Calculate the sum of the following examples: Addition 1. 0.251 + 10.298
Multiplication 1. 5.05 x 13.8
2. 18.098 + 210.099
2. 1.27 x 0.871
3. 0.025 + 10.995
3. −1.01 x 0.89
4. 1.09 x 104 + 1.2 x 102
4. 27.3 x −9.31
5. 27.3 + 0.021 + 68.3
5. 1.09 x 104 x 1.2 x 102
Subtraction 1. 27.3 − 4.36
Division 1. 233.1
2. 21.76 − 18.51
2. 0.1254 0.057
3. 32.76 − 20.086
3. 0.6875 22
4. 10.75 − 19.999 − 21.100
4. 24.024 4.62 ’ 5. 1.09 x 104 12
5. 1.09 x 104 − 1.2 x 102
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18.5
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1.6 Signed Numbers 1.6.1 Adding Signed Numbers
1.6.3 Multiplying Signed Numbers
When adding two or more numbers with the same sign, ignore the sign and find the sum of the values and then place the common sign in front of the answer. In other words, adding two or more positive numbers always results in a positive sum, where as adding two or more negative numbers results in a negative sum. When adding a positive and negative number, find the difference between the two numbers and apply (+ or −) of the larger number. In other words, adding negative number is the same as subtracting a positive number. The result of adding or subtracting signed numbers is called algebraic sum of those numbers.
Multiplication of signed numbers is accomplished in the same manner as multiplication of any other number. However, after multiplying, the product must be given a sign. There are three rules to follow when determining a products sign. 1. The product of two positive numbers is always positive. 2. The product of two negative numbers is always positive. 3. The product of a positive and a negative number is always negative.
Add 25 + (−15) 25 + (−15) 10
or
25 − 15 10
1.6.2 Subtracting Signed Numbers When subtracting numbers with different signs, change the operation sign to plus and change the sign of the subtrahend. Once this is done, proceed as you do in addition. For example +3 − −4 is the same as +3 + +4. There is no difference if the subtrahend is larger than the minuend, since the operation is done as though the two quantities are added.
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Example: Subtract 48 from −216 Step 1: Set up the subtraction problem −216 − 48 Step 2: Change the operation sign to a plus sign and change the sign of the subtrahend. Now add. −216 + −48 = −264
Example: −6 x −2 = 12 6 x 2 = 12 (−6) x (−2) = 12 (−6) x 2 = −12
1.6.4 Dividing Signed Numbers Like multiplying signed numbers, division of signed numbers is accomplished in the same manner as dividing any other number. The sign of the quotient is determined using the rules identical to those used in multiplication. Example: 12
3=4 12 (−3) = −4 (−12) (−)3 = 4 (−12) 3 = −4
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M1 Calculate the sum of the following examples: 1. −8 + 5
16.11 − 12
2. −7 − 6− 3
17.15
3. 8 − 7 − 15
18.10 − 12
4. −3 + 5 + 7 − 4 − 2 5. 6 + 4 − 3 − 5 − 7 + 2 6. 8 x (−3) 7. (−2) x (−5) x (−6) 8. 4 x (−3) x (−2) 9. (−3) x (−4) x 5 10.−16
(−2) x (−4)
11.15 x (−3) x 2
(−5) x (−6)
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13. 7 x 5 − 2 + 4 x 6 14.7 x 5 − 12 15.11 − 9
4 + 3
3 + 7
4 + 3 x (6 − 2)
(4 + 1) − 9 x 3 + 7 (4 + 3) 6 + 3 (8 − 3)
Question 1. 16 holes spaced 48mm apart are to be marked off on a sheet of metal. 17 mm is to be allowed between the centres of the holes and the edge of the metal. Calculate the total length of metal required. Question 2. In the first 2 hours of a shift an operator makes 32 soldered joints per hour. In the next 3 hours the operator makes 29 joints per hour. In the final two hours 26 joints are made per hour. How many soldered joints are made in the 7 hours. Question 3. A machinist makes 3 parts in 15 minutes. How many parts can he produce in an 8 hour shift allowing 20 minutes for starting and 10 minutes for finishing the shift. Question 4. The length of a metal plate is 891 mm. Rivets are placed 45 mm apart and the distance between the centres of the end rivets and the edge of the plate is 18mm. How many rivets are required. Question 5. 32 pins each 61 mm long are to be turned in a lathe. If 2 mm is allowed on each pin for parting off. what total length of material is required to make the pins.
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1.7 Common Fractions 1.7.1 Introduction
1.7.2 Lowest Terms
A common fraction represents a portion or part of a quantity. For example, if a number is divided into three equal parts, each part is one−third ( ) of the number. A fraction consists of two numbers, one above and one below a line, orfraction bar. The fraction bar indicates division of the top number, or numerator, by the bottom number, or denominator. For example, the fraction indicates that three is divided by four to find the decimal equivalent of 0.75. When a fractions numerator is smaller than the denominator, the fraction is called a proper fraction. A proper fraction is always less than 1. If the numerator is larger than the denominator, the fraction is called an improper fraction. In this situation the fraction is greater than 1. If the numerator and the denominator are identical, the fraction is equal to 1. A mixed number is the combination of a whole number and a proper fraction. Mixed numbers are expressed as 1 and 29 and are typically used in place of improper fractions. The numerator and denominator of a fraction can be changed without changing the fractions value. A mixed number can be converted into an improper fraction and vice versa.
A fraction is said to be in its lowest terms when it is impossible to find a number which will divide exactly into both its numerator and denominator. The fractions are both in their lowest terms but the fraction is not in its lowest terms because it can be reduced to by dividing top and bottom numbers by 2.
Example:
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Convert 8 = = Express as a mixed number = 6 (since 27 4 = 6 remainder 3)
Example: Reduce to its lowest terms is equivalent to
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M1 Solve the following equations: Convert the following mixed numbers to improper fractions: 1. 2. 3. 4. 5.
Convert the following improper fractions to mixed numbers: 1. 2. L N O S E S O P R U P G N I N I A R T R O F
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3.
4. 5.
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1.7.3 Comparing the Size of Fractions
1.7.4 Adding & Subtracting Fractions
When the values of two or more fractions are to be compared, express each of the fractions with the same denominator. This common denominator should be the LCM of the denominator of the fractions to be compared. It is sometimes called the lowest common denominator (LCM). Example: Arrange the fractions in order of size beginning with the smallest. The LCM of the denominators 6, 8, and 9 is 72, i.e. the lowest common denominator is 72.
Two fractions which have the same denominator can be added together by adding their numerators. Thus
is equivalent to is equivalent to is equivalent to Because all the fractions have been expressed with the same denominator all that we need to do is to compare the numerators. Therfore the order of size is
When two fractions have different denominators they cannot be added together directly. However, if we express the fractions with the same denominator they can be added. Example: Add The lowest common denominator of 5 and 7 is 35
=
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=
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M1 When mixed numbers are to be added together, the whole numbers and the fractions are added separately.
When mixed numbers are involved first subtract the whole numbers and then deal with the fractional parts. Example:
Example: Add
= = = = Two fractions to be subtracted which do not do have the same denominator, a method similar to that for addition is used.
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Example: Subtract from The lowest common denominator is 12
= =
Subtract = = =
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1.7.5 Multiplication of Fractions Multiplication of fractions is performed by multiplication the numerators of each fraction to form the product numerators, and multiplying the individual denominators to form the product denominator. The resulting fraction is then reduced to its lowest terms. Example: Multiply
= If any factors are common to a numerator and a denominator they should be cancelled before multiplying. Example:
Find the value of = L N O S E S O P R U P G N I N I A R T R O F
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Mixed numbers must be converted into improper fractions before multiplying. Example: Multiply
= = = In problems with fractions the word “of” is frequently used. It should always be taken as meaning “multiply”.
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1.7.6 Division of Fractions Division of common fractions is accomplished by inverting, or turning over, the divisor and then multiplying. However, it is important that you invert the divisor only and not the dividend. Once the divisor is inverted, multiply the numerators to obtain a new numerator, multiply the denominators to obtain a new denominator, and reduce the quotient to its lowest terms. Example: Divide
= = = Mixed numbers must be converted into improper fractions before multiplying.
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Solve the following equations: Add the following fractions:
Multiply and simplify the following fractions:
1.
1.
2.
2.
3.
3.
4.
4.
5.
5. of 16
Divide and simplify the following fractions: Subtract the following fractions:
1. 2. L N O S E S O P R U P G N I N I A R T R O F
3. 4. 5.
1. 2. 3.
4.
5.
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M1 Arrange the following sets of fractions in order of size:
1. 2.
3.
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1.8 Decimals Working with fractions is typically time consuming and complex. One way you can eliminate fractions in complex equations is by replacing them with decimal fractions or decimals. A common fraction is converted to a decimal fraction by dividing the numerator by the denominator. For example, is converted to a decimal by dividing the 3 by the 4. The decimal equivalent of is 0.75. Improper fractions are converted to decimals in the same manner. However, whole numbers appear to the left of the decimal point. In a decimal, each digit represents a multiple of ten. The first digit represents tenths, the second hundredths, the third thousandths. Example: 0.5 is read as five tenths 0.05 is read as five hundredths 0.005 is read as five thousandths when writing decimals, the number of zeros to the right of the decimal does not affect the value as long as no other number except zero appears. In other words, numerically, 2.5, 2.50 and 2.500 are the same. The number of digits after the decimal point are called decimal places Examples: 27.6 one decimal point 27.16 two decimal points 27.026 three decimal points and so on.
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1.8.1 Adding Decimals The addition of decimals is done in the same manner as the addition of whole numbers. However, care must be taken to correctly align the decimal points vertically. Example: Add the following 25.78 + 5.4 + 0.237 rewrite with the decimals aligned and add. 25.78 5.4 + 0.237 31.417 Once everything is added, the decimal point in the answer is placed directly below the other decimal points.
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1.8.2 Subtracting Decimals
1.8.4 Dividing Decimals
Like adding, subtracting decimals is done in the same manner as with whole numbers. Again, it is important that you keep the decimal points aligned.
When dividing decimals, the operation is carried out in the same manner as division of whole numbers. However, to ensure accurate placement of decimal point in the quotient, two rules apply:
Example: If you have 325.25 pounds of ballast on board and remove 30.75 pounds, how much ballast remains? 325.25
− 30.75 294.50
1. When the divisor is a whole number, the decimal point in the quotient aligns vertically with the decimal in the dividend when doing long division. 2. When the divisor is a decimal fraction, it should first be converted to a whole number by moving the decimal point to the right. However, when the decimal in the divisor is moved, the decimal in the dividend must also move in the same direction and the same number of spaces.
1.8.3 Multiplying Decimals When multiplying decimals, ignore the decimal points and multiply the resulting whole numbers. Once the product is calculated, count the number of digits to the right of the decimal point in both the multiplier and multiplicand. This number represents the number of places from the left the decimal point is placed in the product.
Example: Divide 37.26 by 2.7 Move the decimal in the divisor to the right to convert it to a whole number. 27 ) 37.26
Example: 26.757 L N O S E S O P R U P G N I N I A R T R O F
3 decimal x 0.32 2 decimal 53514 80271 856224 count 5 decimal places to the left of the digit 4 8.56224
Move the decimal in the dividend the same number of places to the right. 27 ) 372.6 Divide:
13.8 27 ) 372.6 27 102 81 216
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1.9 Conversions 1.9.1 Converting Decimals to Fractions Although decimals are typically easier to work with, there are times when the use of a fraction is more practical. For example, when measuring something, most scales are fractional increments. For this reason it is important that you know how to convert a decimal number into a fraction. For example, 0.125 is read as 125 thousandths, which is written as 125/1000. This fraction is then reduced to its lowest terms.
Convert
to decimals
= 0.84375
To convert a fraction into a decimal we divide the denominator into the numerator.
Examples:
1.9.2 Converting Fractions to Decimals
When we have mixed numbers to convert into decimals we need only deal with the fractional part. Thus to convert into decimals we only have to deal with
= 0.5625 The division shows that and hence = 2.5625. Sometimes a fraction will not divide out exactly. If the number is recurring the answer can be given to 1 or 2 decimal places or that specified by the equation.
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1.9.3 Converting Fractions to Percentages To change a fraction to a percentage you must multiply by 100. Example:
as a percentage = = = 60% as a percentage = = = 475%
1.9.4 Converting Percentages to Fractions To change a percentage to a fraction, divide by 100%. Examples: 8% as a fraction =
12 % (12.5) as a fraction =
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1.9.5 Converting Percentages to Decimals
To convert a percentage to a decimal, firstly, convert the percentage to a fraction, then the fraction to a decimal. Examples: 65% as a fraction =
, as a decimal = 0.65
32 % as a fraction = , as a decimal = 0.325
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1.9.6 Converting Decimal to a Percentage To convert a decimal to a percentage, firstly, convert the decimal to a fraction, then convert the fraction to a percentage.
1.9.6.2
Example:
To express one quantity as a percentage of another, make a fraction of the two quantities and multiply by 100.
0.021 as a fraction = , as a decimal = 2.1%
Expressing one Quantity as a Percentage
Example:
, as a decimal = 3.7% 0.037 as a fraction =
12 as a percentage of 50 = = 24%
0.43 as a fraction =
4 as a percentage of 60 = = 6.67%
1.9.6.1
, as a decimal = 43%
Values of a Percentage of a Quantity
To find the value of a percentage of a quantity, firstly, express the percentage as a fraction and multiply by the quantity.
Examples: 4% of 60 = = = = 3 % of 1500 = L N O S E S O P R U P G N I N I A R T R O F
=
3.2 as a percentage of 2.4 = = 13.333% or %
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M1 Solve the following equations: Convert the following decimals to fractions in their lowest terms: 1. 0.2 2. 0.45 3. 0.3125 4. 2.55 5. 0.0075 6. 2.125 Find the difference between: 1. and 0.295 2. and 1.1632 Convert the following fractions to decimals (3 decimal places): 1. 2. 3. 4. 5.
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Place the following in ascending order of size: 1. 2. 3.
Express the following as a percentage %: 1. 0.43 2. 0.025 3. 1.25 4. 5. 6. 7. Express the following as fractions: 1. 25% 2. 13% 3. 4.5% 4. % 5. 33% Express: 1. 2. 3. 4. 5.
30 as a percentage of 50 24 as a percentage of 16 0.5 as a percentage of 12.5 3.2 as a percentage of 2.4 0.08 as a percentage of 0.72
Calculate: 1. 4% of 30 2. 0.8% of 360 3. 1.5% of 60 4. 120% of 75 5. 80% of 90
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1.10 Ratio & Proportion 1.10.1
Ratio
A ratio is a comparison between two similar quantities. If the length of an aircraft is 75 m and a model of it is 1 m long then the length of the model is of the length of the aircraft. In making the model all the dimensions of the aircraft are reduced in the ratio of 1 to 75. The ratio 1 to 75 is usually written 1 : 75. A ratio can also be written as a fraction, as indicated above, and a ratio of 1:75 means the same as the fraction . Before we can state a ratio the units must be the same. we can state a ratio between 3 mm and 2 m provided we bring both lengths to the same units. Thus if we convert 2 m to 2000 mm the ratio between lengths is 3:2000. Example: Express the following ratios as fractions reduced to their lowest terms: i. 40 mm to 2.2 m 2.2 m = 2200 40 : 2200 = =
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ii. 800 g to 1.66 kg 1.6 kg = 1600 g 800 : 1600 = =
A ratio provides a means of comparing one number to another. For example, if an engine turns at 4,000 rpm and the propeller turns at 2,400 rpm, the ratio of the two speeds is 4,000 to 2,400, or 5 to 3, when reduced to lowest terms. This relationship can also be expressed as 5/3 or 5:3. The use of ratios is common in aviation. One ratio you must be familiar with is compression ratio, which is the ratio of cylinder displacement when the piston is at bottom centre to the cylinder displacement when the piston is at top centre. For example, if the volume of a cylinder with the piston at bottom centre is 96 cubic inches and the volume with the piston at top centre is 12 cubic inches, the compression ratio is 96 :12 or 8 :1 when simplified. Another typical ratio is that of different gear sizes, for example, the gear ratio of a drive gear with 15 teeth to a driven gear with 45 teeth is 15:45 or 1:3 when reduced. This means that for every one tooth on the drive gear there are three teeth on the driven gear. However, when working with gears, the ratio of teeth is opposite the ratio of revolutions. In other words, since the drive gear has one third as many teeth as the driven gear, the drive gear must complete three revolutions to turn the driven gear one revolution. This results in a revolution ratio of 3:1, which is opposite the ratio of teeth. Before we can state a ratio the units must be the same. We state a ratio between 3 mm and 2 m provided we bring both lengths to the same units. If convert 2 m to 2000 mm the ratio between the two lengths is 3:2000.
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1.10.2
Proportion
1.10.2.1 General
1.10.2.2 Direct Proportion
A proportion is a statement of equality between two or more ratios and represents a convenient way to solve problems involving ratios. For example, if an an engine has a reduction gear ratio between the crankshaft and the propeller of 3:2 and the engine is turning 2,700 rpm, what is the speed of the propeller? In this problem, let “X” represent the unknown value, which in this case is the speed of the propeller. Next set up a proportional statement using the fractional form .
If 5 litres of oil has a mass of 4 kg, then 10 litres of the same oil will have a mass of 8 kg. That is, if we double the quantity of oil its mass is also doubled. Now 2 litres of oil will have a mass of 2 kg. That is if we halve the quantity of oil we halve its mass. This is an example of direct proportion. As the quantity of oil increases the mass increases in the same proportion. As the quantity of oil decreases the mass decreases in the same proportion.
To solve this equatiuon, cross multiply to arrive at the equation 3x = 2 x 2,700, or 5,400. to solve for (x), divide 5,400 by 3. The speed of the propeller is 1,800 rpm.
Thus resistance of wire 1m long = = 13.3 ohms
1.10.2.3 Inverse Proportion
This same proportion may also be expressed as 3:2 = 2,700 : X. The first and last terms of the proportion are called extremes, and the second and third terms are called the means. In any proportion, the product of the extremes is equal to the product of the means. In this example, multiply the extremes to get 3x, and multiply the means to get 2 x 2,700 or 5,400. This results in the identical derived earlier; 3x = 5,400. L N O S E S O P R U P G N I N I A R T R O F
3:2 = engine speed : propeller speed 3x = 2 : 2,700 3x = 5,400 x = 1,800 rpm.
Example: The electrical resistance of a wire 150 mm long is 2 ohms. Find the resistance of a similar wire which is 1 m long. The lengths of the two wires are increased in the ratio of 1000 :150. The resistance will aslo increase in the ratio 1000 :150.
3:2 = 2,700 : x
A motor car will travel 30 km in 1 hour if its speed is 30 km per hour. If its speed is increased to 60 km per hour the time taken to travel 30 km will be hour. That is when the speed is doubled the time taken is halved. This is an example of inverse proportion. When we multiply the speed by 2 we divided the time taken by 2. Example: Two pulleys of 150 mm and 50 mm diameter are connected by a belt. If the larger pulley revolves at 80 rev/min find the speed of the smaller pulley. The smaller pulley must revolve faster than the larger pulley and hence the quantities, speed and diameter, are in inverse proportion. The pulley diameters are decreased in the 50 : 150, or 1 : 3. The speed will be increased in the ratio of 3 :1. Therefore Speed of smaller pulley = 240 rev/min.
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1.10.2.4 Proportional Parts The diagram below shows the line AB whose length represents 10 m divided into two parts in the ratio 2:3. From the diagram the line has been divided into a total og 5 parts. The length AC contains 2 parts and the length BC contains 3 parts. Each part is 2m long, hence AC is 4m long and BC is 6m long.
Example: A certain brass is made by alloying copper and zinc in the ratio of 7:3. How much copper must be mixed with 30 g of zinc. 3 parts have a mass of 30 g 1 part has a mass of 10 g 7 parts have a mass of 70 g Therefore,
We could tackle the problem as follows; Total number of parts = 2 + 3 = 5 Length of each part
L N O S E S O P R U P G N I N I A R T R O F
= = 2 m
Length of AC = 2 x 2 = 4 m Length of BC = 3 x 2 = 6 m
Mass of copper needed = 70 g.
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M1 Solve the following equations: Express the following ratios as fractions reduced to their lowest terms: 1. 15 g to 2 kg 2. 30 p to £ 5 3. 20 cm to 100 mm 4. 400 m to 3 km 5. 21 ft to 9 inches Find the missing value: 1. 3 : 4 = 6 : x 2. 20 : 1 = x : 3.2 3. 240 : 400 = x : 1 4. 1 : 2.6 = x : 13 5. 18 : x = 2 : 1 Five men build a wall take 20 days to complete it. How long would it take 4 men to complete it. 4 people can clean an office in 6 hours. How many people would be needed to clean the office in 4 hours. 8 people take 5 hours to change an engine. How long would it take 4 people to do this work.
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An engineering company employ 12 men to fabricate a number of containers. They take 9 days to complete the work. If the company had employed 8 men, how long would it have taken. A train travels 200 km in 4 hours. If it travels at the same rate, how long will it take to complete a journey of 350 km. A motor running at 400 rev/min has a pulley of 125 mm diameter attached to its shaft. It drives a parallel shaft which has a 1000 mm diameter pulley attached to it. Find the speed of this shaft.
A gear wheel having 40 teeth revolves at 120 rev/min. It meshes with a wheel having 25 teeth. Find the speed of the 25 tooth wheel. Two shafts are to rotate at 150 and 250 rev/min respectively. A 120 mm diameter pulley is fitted to the slower shaft and by means of a belt it drives a pulley on the faster shaft. What diameter pulley is required on the faster shaft. A bar of metal 10.5 m long is to be cut into three parts in the ratio of . Find the length of each part.
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1.11 Power and Roots When a number is multiplied by itself, it is said to be raised to a given power. For example, 6 x 6 = 36; therefore, = 36. The number of times a base number is multiplied by itself is expressed as an exponent and is written to right and slightly above the base number. A positive exponent indicates how many times a number is multiplied by itself. Example:
is read 3 squared or 3 to the second power. its value is found by multiplying 3 by itself 2 times. 3x3=9
is read 2 cubed or 2 to the third power. Its value is found by multiplying 2 by itself 3 times. 2 x 2 x2 = 8 A negative exponent implies division or fraction of a number. It indicates the inverse, or reciprocal of the number with its exponent made positive.
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Example: 2−3 is read 2 to the negative third power. the inverse, or reciprocal of 2−3 with its exponent made positive is
Any number, except zero, that is raised to the zero power equals 1. When a number is written without an exponent, the value of the exponent does not have a SIGN (+ or −) preceding it, the exponent is assumed to be positive. The root of a number is that value which, when multiplied by itself a certain number of times, produces that number. For example, 4 is a root of 16 because when multiplied by itself, the product is 16. However, 4 is also a root of 64 because 4 x 4 x 4 = 64. The symbol used to indicate a root is the radical sign ( ) placed over the number. If only the radical sign appears over a number, it indicates you are to extract the square root of the number under the sign. The square root of a number is the root other than a square root, an index number is placed outside the radical sign. for example the cube root is expressed as
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1.12 Indices 1.12.1
Base, Index & Power
The quantity may be written as . Now is called the fourth power of the base 2. The number 4, which gives the number of 2s to be multiplied together is called the index (plural : Indices). Similarly
Here is the third power of the base , and the index is 3. Thus in this expression n n is called the nth power of is called the base, and n is called the index. Remember that, in algebra, letters such as in the above expression merely represent numbers. Hence the laws of arithmetic apply strictly to algebraic terms as well as numbers.The expression is called the reciprocal of 2. is called the reciprocal of likewise the expression Similarly the expression is called the reciprocal of
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Solve the following equations: Find the values of the following: 1. 2. 3. 4.
6. 7. 8. 9. 10. 5.
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1.12.2
Laws of Indices
1.12.2.1 Multiplication
1.12.2.2 Division of Powers
Let us see what happens when we multiply powers of the same base together. = or
Now lets see what happens when we divide powers of the same base.
= In both the examples above we see that we could have obtained the result by adding the indices together. We may apply this idea when multiplying more than two powers of the same base together.
Thus The law is:
When multiplying powers of the same base together, add the indices.
We see that the same result could have been obtained by subtracting the indices. The law is:
When dividing powers of the same base subtract the index of the denominator from the index of the numerator. 1.12.2.3 Powers of Powers
How do we simplify ? One way is to proceed as follows:
53+3 = We that the same result would have been obtained if we multiply the two indices together.
53x2 = 56 The law is: L N O S E S O P R U P G N I N I A R T R O F
When raising the power of a base to a power, multiply the indices together.
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1.12.2.4 Zero Index
1.12.2.6 Fractional Indices
Now
The cube root of 5 (written as ) is the number which, when multiplied by itself three times, gives 5.
But using the laws of indices
Thus or = 1 Also The law is:
Any base raised to the index zero is equal to 1.
but we also know that Comparing these expressions
Similarly the fourth root of base ( written as ) is the number which, when multiplied by itself four times, gives .
But we also know that
1.12.2.5 Negative Indices but using the Now laws of indices it follows that The law is:
The power of a base which has a negative index is the reciprocal of the power of the base having the same, but positive, index.
L N O S E S O P R U P G N I N I A R T R O F
Comparing these expressions The law is:
A fractional index represents a root, the denominator of the index denotes the root to be taken.
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M1 Solve the following equations: Simplify the following, giving each answer as a power
Find the value of the following
1.
1. 811/4
2.
2. 82/3
3.
3. 16−3/4
4. 5. 6. 7.
8.
9.
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4. 92.5
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1.13 Tranposition of Formulae The formula has as its subject. By rearranging this formula we could make the subject. We are then said to have transposed the formula to make the subject. The rules for transforming a formula are: 1. Remove square roots or other roots. 2. Get rid of fractions. 3. Clear brackets. 4. Collect together the terms containing the required subject. 5. Factorise if necessary.
to make the subject. Step 1. Since there are no roots get rid of the fraction by multiplying both sides of the equation by . iii. Transpose the formula
Step 2. Clear the bracket.
6. Isolate the required subject. These steps should be performed in the order given.
Step 3. Collect the terms containing on the LHS.
Examples: i. Transpose the formula F = ma to make a the subject. Step 1. Divide both sides by m. then,
Step 4. Factorise the LHS.
or
or
to make b the subject Step 1. Multiply both sides by b. then, or ii. Transpose
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Step 5. Isolate by dividing both sides of the equation by . Although we used five steps to obtain the required subject, in very many cases far fewer steps are needed. Nevertheless, you should work through the steps in the order given. iv. Tranpose to make the subject. Step 1. Remove the square root by squaring both sides. Step 2. Since there are no fractions or brackets and factorisation is not needed we can now isolate by dividing both sides of the equation by . or Since it is usual to position the subject on the LHS.
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M1 Transpose the following: 1. for d
15.
for N
16.
2. for d 3. for R 4. for h 5. for y 6. for T 7. for t 8. 9.
for R
for J
for t
11.
for r
12.
for x
13.
10.
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14.
for x
for E
for t
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1.14 Areas The area of a plane figure is measured by seeing how many square units it contains. 1 square metre is the area contained in a square metre is the area contained in a square having a side of 1 metre; 1 square centimetre is the area contained in a square having a side of 1 centimetre, etc. The standard abbreviations are 1 square metre 1 m2 1 square centimetre 1 cm2 1 square millimetre 1 mm2 1 square inch 1 square foot 1 square yard
1 in2 1 ft2 1 yd2
The following provides the formulae for areas and perimeters of simple geometrical shapes. Rectangle
L N O S E S O P R U P G N I N I A R T R O F
Area = Perimeter =
Parallelogram
Area = Perimeter = Sum of all 4 sides.
Triangle
Area = or Area = , where
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Solve the following problems: 1. Find the area and perimeter of a rectangle whose length is 12 inches and width is 7 inches. 2. A rectangle lawn is 35 m long and 18 m wide. A path 1 m wide is made round the lawn. Calculate the area of the path. 3. A carpet has an area of 36 m2. If it is square what length of side has the carpet. 4. A triangle has a base of 7 cm amd an altitude of 3 cm. Calculate its area. 5. A triangle has sides which are 8 cm , 12 cm and 14 cm long. Determine the area ofthe triangle. 6. The area of a triangle is 40 ft2. Its base is 8 ft long. Calculate its vertical height.
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Trapezium Sector of a Circle
Area =
Area =
Circle
L N O S E S O P R U P G N I N I A R T R O F
Area = Circumference =
or
Perimeter =
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Solve the following problems: 7. Find the area of a trapezium whose parallel sides are 45 mm and 73 mm long if the distance between them is 42 mm. 8. An annulus has an outer diameter of 10 cm and an inner diameter of 6 cm. calculate its area. 9. Find the area and length of arc of the sector of a circle, with an angle of 150 and a radius of 21 mm.
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1.15 Volumes The volume of a solid figure is found by seeing how many cubic units it contains. 1 cubic metre is the volume contained inside a cube having an edge 1 metre long; 1 cubic centimetre is the volume contained inside a cube having an edge 1 centimetre long, etc. The standard abbreviations for units of volume are as follows: 1 square metre 1 square centimetre
1 m3 1 cm3
1 square millimetre 1 square inch 1 square foot 1 square yard
1 mm3 1 in3 1 ft3 1 yd3
Cylinder
Volume = Surface Area =
The following figures give the formulae for the volumes and surface areas of solid figures. Any solid having a uniform cross−section;
Cone
Volume = Cross−sectional area x Length of solid Surface Area = Lateral Surface + Ends i.e. (perimeter of cross−section x Length of Solid) + (Total area of ends) L N O S E S O P R U P G N I N I A R T R O F
Volume = ( h is the vertical height) Surface Area = ( l is the slant height)
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Solve the following problems: 10.Calculate the volume of a metal pipe whose inside diameter is 6 cm and whose outside diameter is 8 cm, if it 20 cm long. 11.A sphere has a diameter of 8 in. Calculate its volume and surface area. 12.Find the volume of a cone with a diameter of 7 cm and vertical height of 4 cm.
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Frustum of a Cone Pyramid
Volume = Curved Surface Area =
( h is the vertical height)
Total Surface Area = ( l is the slant height)
Sphere L N O S E S O P R U P G N I N I A R T R O F
Volume =
Volume = Surface Area = Sum of the areas of the triangles forming the sides plus the area of the base ( A = Area of base)
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Solve the following problems: 13.A pyramid has a square base of side 12 cm and an altitude of 15 cm. Calculate its volume. 14. A rectangle tank is 2.7 m long, 1.8 mm wide and 3.2 high. How many litres of water will it hold when full.
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1.16 Conversions Length 1 in 1m 1 ft 12 in 3 ft 1 yd 1 km 1 mile
= = = = = = = =
Mass 2.54 cm 39.37 in or 3.281 ft 0.3048 m 1 ft 1 yd 0.9144 m 0.621 miles 1.61 km = 5,280 ft
1 amu 1000 kg 1000 g 1 slug
= = = =
1.66 x 10−27 kg 1t 1 kg 14.59 kg
Force & Weight 1N 1N
= =
0.2248 lbf 3.5969 ozf
Area m2
1 1 m2 1 ft2
= = =
10.76 ft2 10.000 cm2 0.0929 m2 = 144 in2
1 in2
=
6.452 cm2
Volume
L N O S E S O P R U P G N I N I A R T R O F
1 m3 1 ft3 1 litre 1 ft3 1 gal 1 gal
= = = = = =
1.000.000 cm3 1728 in3 = 0.0283 m3 1000 cm3 = 1.0576 qt 7.481 gal 8 pints 4.546 litres = (3.785 litres in American)
Velocity 1 mph 1 m/s 1 knot 1 knot 1 knot 1 mph
= = = = = =
1.47 ft/s 3.281 ft/s 1.688 ft/s 1.151 mph 1.852 km/h 1.61 km/h
Energy 1J 1 cal 1 Btu
= = =
0.738 ft lb 4.186 J 252 cal
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Time 1 year 1 day
= =
365 days 24 h = 1,440 min
Power 1 HP 1 HP 1W 1W
= = = =
550 ft lb/s 746 W 1 J/s 0.738 ft lb/s
1 Btu/h
=
0.293 W
Pressure 1 atm 1 atm 1 atm 1 atm 1 Pa 1 bar
= = = = = =
76.0 cmHg 760 mmHg 29.92 inHg 14.7 lb/in2 0.000145 lb/in2 14.5 lb/in2
1 bar
=
100,000 Pa
Other Useful Data L N O S E S O P R U P G N I N I A R T R O F
1 litre water = 1 pint water =
1 kg 1 lb
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M1 Convert the following weights and measures: 1. Convert 6 m to feet. 2. What is 4 ft 6in to cm. 3. Convert 25 US gallons to litres. 4. Convert 254 inches to cm. 5. Convert 4.5 litre to US gallons 6. Convert 350 imperial gallons to litres. 7. What is 220 lb in kilos. 8. What is 220 kilos in litres. 9. How many pounds in 1 metric tonne. 10.Convert the following to F −20 C −5 C 37 C 88 C
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11.Convert the following to C −40 F 16 F 100 F 215 F
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1.17 Test Work out the value of the following: 1. 7 + 4 x 3 =
14.Remove the backets from the following (2x + 5)(x + 3) =
2. 5 x 4 − 3 x 6 + 5 =
15.Solve the equation 7x + 3 = 5x + 17 for the value of x.
3. 10 − 12
6
+ 3 (8 − 3) =
4. 53 = 5. = 6.
=
7. = 8. = 9. divide 74.52 by 8.1 = 10.multiply 20.3 x 17.4 = L N O S E S O P R U P G N I N I A R T R O F
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11.Convert 0.800 to a fraction = 12.convert to a percentage = 13.Transpose the following formula to find (u) v = u + at
16.A triangle has length of sides 3 cm and 4 cm. Using pythagros theorem calculate the missing length.
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2.
ALGEBRA
2.1 Introduction
2.2 Use of Symbols
The methods of algebra are an extension of those used in arithmetic. In algebra we use letters and symbols as well as numbers to represent quantities. When we write that a sum of money is £ 50 we are making a particular statement but if we write a sum of money is £ P we are making a general statement. This general statement will cover any number we care to substitute for P.
A technician often has to indicate that certain quantities or measusrements have to be added, subtracted, multiplied or divided. Frequently this has to be done without using actual numbers. The statement: Area of a rectangle = length x breadth is a perfectly general statement which applies to all rectangles. If we use symbols we obtain a much shorter statement. if and
A = the area of the rectangle l = the length of the rectangle b = the breadth of the rectangle
then the statement becomes: A = l x b Knowing what the symbols A, l and b stand for, this statement conveys as much information as the first statement. To find the area of a particular rectangle we replace the symbols l and b by the actual dimensions of the rectangle, first making sure that l and b have the same units. To find the area of a rectangle whose length is 50mm and whose breadth is 30 mm we put l = 50 mm and b = 30 mm. A = l x b = 50 x 30 = 1500 mm2 L N O S E S O P R U P G N I N I A R T R O F
Many verbal statements can be translated into symbols as the following statements show: The difference of two numbers = x − y Two numbers multiplied together = a x b One number divided by another = p q
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2.3 Substitution The process of finding the numerical value of an algebraic expression for given values of the symbols that appear in it is called substitution. Example: If x = 3, y = 4 and z = 5 find the value of:
=
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= = 2.75 or 2 .
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M1 Solve the following substitution equations: If a = 2, b = 3 and c = 5. Find the values of the following. 1. a +7 2. 9c 3. 3bc 4. 4c + 6b 5. a + 2b + 5c 6. 8c − 4b 7.
8.
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2.4 Addition & Substraction of Algebraic Terms Like terms are numerical multiples of the same algebraic quantity. 7x, 5x and −3x are three like terms. An expression consiting of like terms can be reduced to a single term by adding or substracting the numerical coefficients. 7x − 5x + 3x = (7 − 5 + 3) x = 5x 3b2 + 7b2 =
(3 + 7)
b2 =
10b2
−3y − 5y = ( −3 −5) y = −8y q − 3q = (1 − 3) q = −2q Only like terms can be added or subtracted. Thus 7a + 3b − 2c is an expression containing three unlike terms and it cannot be simplified any further. Similarly with 8a2b + 7ab3 − 6a2b2 which are all unlike terms. It is possible to have several sets of like terms in an expression and each set can then be simplified. L N O S E S O P R U P G N I N I A R T R O F
8x + 3y − 4z − 5x + 7z − 2y + 2z = (8 − 5)x + (3 − 2)y + (−4 +7 + 2)z = 3x + y + 5z
2.5 Multiplication & Division Signs When using symbols multiplication signs are nearly always omitted and l x b becomes lb. Of course the same scheme cannot apply to numbers and we cannot write 9 x 6 as 96. The multiplication sign can, however, be omitted when a symbol and a number are to be multiplied together. Thus 5 x m is written 5m. The system may be extended to three or more quantities and hence P x L x A x N is written PLAN. The symbols need not be written in any special order because the order in which numbers are multiplied together is unimportant. Thus PLAN is the same as LANP or NAPL. It is usual, however, to write numbers before symbols, that is, it is better to write 8xy than xy8 or x8y. In algebraic expressions the number in front of the symbols is called the coefficient. Thus in the expression 8x the coefficient of x is 8. The division sign P
is seldom used in algebra and it is more convenient to write
q in the fractional form
Example:
=
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M1 Simplify the following: 1. 7x + 11x
1. 2z x 5y
2. 7x − 5x
2. 3a x 3b
3. 3x − 6x
3. 3 x 4m
4. −2x −4x
4. q x 16p
5. −8x + 3x
5. z x (y)
6. −2x + 7x
6. a x a
7. 5m + 13m − 6m
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M1 Simplify the following: 1. 6b2 − 4b2 + 3b2 2. 6ab − 3ab − 2ab 3. 14xy + 5xy − 7xy + 2xy 4. −5x + 7x − 3x − 2x 5. 3x − 2y + 4z − 2x − 3y + 5z + 6x + 2y − 3z
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2.6 Muliplication & Division of Algebraic Quantities The rules are exactly the same as those used with directed numbers.
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M1 Simplify the following: 1. (−3a) x (−2b) 2. 8m x (−3n) 3. (−4a) x 3b 4. 8p x (−q) x (−3r) 5. 3a x (−4b) x (−c) x 5d 6. 3m x (−3m)
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When multiplying expressions containing the same symbols, indices are used:
Remember the word BODMAS which gives the initial letters of the correct sequence i.e. Brackets, Of, Division, Multiply, Add, Subtract. Thus
=
= =
When dividing algebraic expressions, cancellation between numerator and denominator is often possible, cancelling is equivalent to dividing both numerator and denominator by the same quantity:
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M1 Simplify the following: 1. 4ab
2a
2. 12x2yz2
4xz2
3. (−12a2b) 4. 8a2bc2 5. 7a2b2
6a
4ac2
3ab
6. 8mn x (−3m2n3) 7. 7ab x (−3a2) 8. m2n x (−mn) x 5m2n2 9. 5a2 x (−3b) x 5ab
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2.7 Brackets Brackets are used for convenience in grouping terms together. When removing brackets each term within the bracket is multiplied by the quantity outside the bracket:
When simplifying expressions containing brackets first remove the brackets and then add the like terms together.
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Remove the brackets in the following:
Find the products of the following:
1. 3(x 3(x + 4)
1. (x + 4) (x (x + 5) 5)
2. 2(a 2(a + b)
2. (2x + 5) (x + 3)
3. 3(3 3(3x − 2y)
3. (5x + 1) 1) (2x (2x + 3)
4. (x − 1)
4. (7x + 2) 2) (3x (3x + 2)
5. 5(2p 5(2p − 3q)
5. (x − 4) (x − 2)
6. 7(a − 3m)
6. (2x − 1) (x − 4)
7. −(a + b)
7. (2x − 4) (3x − 2)
8. −(a − 2b)
8. (x − 2) (x + 7)
9. −(3p − 3q)
9. (2x (2x + 5) (x − 2) 10.(3x + 4y) (2x − 3y) 11. (2x + 3) 2
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M1 Remove the brackets in the following: 1. −4(x + 3)
Remove the brackets in the following: When a bracket has a minus sign in front of it, the signs of all the terms inside the bracket are changed when the bracket bracket is removed. The reason for this rule may be seen from the following examples:
2. −2(2x − 5) 3. −5(4 − 3x) 4. 2k( 2k(k − 5) 5. −3y(3x + 4) 6. 4xy( 4xy(ab ab − ac + d) 7. 3x2(x2 − 2xy + y2) 8. −7p(2p2 − p + 1)
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M1 Remove the brackets and simplify: 1. 3( x + 1) + 2(x + 4) 2. 5(2a + 4) − 3(4a + 2) 3. 3(x + 4) − (2x + 5) 4. 4(1 − 2x) − 3(3x − 4) 5. 5(2x − y) − 3(x + 2y) 6. (y − 1) + (2y − 3) 7. −(4a + 5b − 3c) − 2(2a − 3b − 4c) 8. 2x(x − 5) − x(x − 2) − 3x(x − 5) 9. 3(a − b) − 2(2a − 3b) + 4(a − 3b)
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2.8 Factorisation A factor is a common part of two or or more terms which make up an algebraic expression. Thus the expression has two terms which have the number 3 common to both of them. Thus . We say that 3 and are the factors of To factorise algebraic expressions of this kind, we first find the Highest Common Factor (HCF) of all the terms making up the expression. The HCF then appears outside the bracket. To find the terms inside the bracket divide each of the terms making up the expression by the HCF.
2.8.1 Factorising by Grouping To factorise the expression ax + ay + bx + by first group the terms in pairs so that each pair of terms has a common factor. Thus; Example: ax + ay + bx + by = (ax + ay) + (bx + by) = a(x + y) + b(x + y) = (a + b) (x + y)
Example: Find the factors of The HCF of and is
Factorise mp + np − mq − nq
mp + np − mq − nq = (mp + np) − (mq − nq) = p(m + n) − q(m + n) = (p − q) (m + n)
Find the factors of The HCF of and is
Find the factors of The HCF of , and is
= L N O S E S O P R U P G N I N I A R T R O F
The HCF of , and is Find the factors of
=
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M1 Factorise each of the following: 1. 5x + 5y 2. 5p − 5q 3. 6x + 18y 4. ax − y 5. 4p − 12q 6. 4x − 6xy 7. 8x2 − 4x 8. ax2 − bx 9. x(a − b) + 2(a − b) 10.m(p + q) − n(p + q) 11.5a − 10b + 15c 12.ab + ac − bd − cd
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13.2pr − 4ps + qr − 2qs 14.4ax + 6ay − 4bx − 6by 15.3mx + 2nx − 3my − 2yn 16.ab(p + q) − cd(p + q) 17.K 2l2 − mnl −k2l + mn
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2.8.2 Factors of Quadratic Expressions An expression of the type is called a quadratic expression.
and
are both quadratic expressions.
In an expression like , the factors will be of the type where and are two numbers whose sum is 7 and whose product is 12. These two numbers must be 3 and 4. So = This techinque is also useful when factorising expressions such as in which , and . We can factorise if we can get two integers whose product is and whose sum is . Thus and . Now . It can now be seen that the two integers are 8 and 15. Since
= = =
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Example:
Factorise Here and . and The two integers whose sum is −34 and whose product is 168 are −6 and −28.
= = =
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M1 Factorise the following: 1. x2 + 5x + 6 2. x2 − 8x + 15 3. x2 − 5x − 6 4. x2 − 5x + 6 5. 2x2 + 7x + 5 6. 2x2 + 13x + 15 7. 3x2 + x − 6 8. 3x2 − 8x + 28 9. 10x2 + 19x − 15 10. 6x2 + x − 35 11.5x2 − 11x + 2 12.6x2 − 7x − 5
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2.8.3 Algebraic Fractions Since algebraic expressions contain symbols (or letters) which represent numbers all the rule of operations with numbers also apply to algebraic terms, including fractions. Thus
and and
You should note in the last example how we put brackets round and to remind us that they must be treated as single expressions, otherwise we may have been tempted to handle the terms and on their own.
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M1 Simplify the following: 1. 2.
3. 4.
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2.8.3.1
Adding & Subtracting Algebraic Fractions
Consider the expression which is the addition of two fractional terms. These are called partial fractions. If we wish to express the sum of these partial fractions as one single fraction then we proceed as follows (similar method used when adding or subtracting number fractions) First find the lowest common denominator. This is the LCM of and which is . each fraction is then expressed with as the denominator. Example: and and adding these new fractions we have:
2.8.3.2
Multiplication & Division of Algebraic Fractions
As with ordinary arithmetic fractions, numerators can be multiplied together, as can denominators, in order to form a single fraction. Example;
or
Factors which are common to both numerator and denominator may be cancelled. it is important to realise that this cancelling means dividing the numerator and denominator by the same quantity. Example:
= =
= = =
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M1 Simplify the following: 1.
13.
14.
2.
3.
4. 5.
6.
7. 8.
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10. 11.
12.
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2.9 Linear Equations An arithmetical quantity has a definite value, such as 93, 3.73 or . An algebraic quantity, however, given by algebraic expressions such as or , represents many amounts depending on the value given to . In algebra there are two methods we use to show that two quantities are equivalent to each other. One is called an identity and the other an equation.
2.9.1 Identities A statement of the type is called an identity The sign means “is identical to”. Any statement using this sign is true for all values of the variable, the variable in this case being .
2.9.2 Equations A statement of the type is called an equation. This means that the quantity on the left−hand side of the equation is equal to the quantity on the right−hand side. We can see that, unlike an identity, there is only one value of that will satisfy the equation, or make the left−hand side equal to the right−hand side. The process of finding is called solving the equation, and the value 8 is known as the solution or root of the equation.
2.9.2.1
Solving Linear Equations
Linear equations contain only the first power of the unkown quantity.
Thus when we have and when we have and so on. Another type of identity involves units as, for example, the relationship between kilometres and metres. This may be stated as km Thus 7 km and 9 km L N O S E S O P R U P G N I N I A R T R O F
m 7000 m 9000 m
and so on.
In practise the sign is often replaced by the = (equals) sign and the above identities would be stated as
and
km = 1000 m
and
are both examples of linear equations. In the process of solving an equation the appearances of the equation may be considerable altered but the values on both sides must remain the same. We must maintain this equality, and hence whatever we do to one side of the equation we must do exactly the same to the other side. After an equation is solved, the solution should be checked by substituting the result in each side of the equation separately. If each side of the equation then has the same value the solution is correct. In the detail which follows, LHS means left−hand side and RHS means right−hand side.
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2.9.2.2
Equation Requiring Multiplication & Division
Example: Solve the equation Multiply each side by 6, we get Check: when , LHS = , RHS = 3
2.9.2.3
Equations Requiring Addition & Subtraction
Example: Solve the equation If we add 4 to each side, we get The operation of adding 4 to each side is the same as transferring −4 to the RHS but in so doing the sign is changed from a minus to a plus. Check: when , LHS = 12 − 4 = 8, RHS = 8
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2.9.2.4
Equations Containing the Unknown Quantity on Both Sides
In equations of this kind, group all the terms containing the unknown quantity on one side of the equation and the remaining terms on the other side. Example; Solve the equation Subtracting 5 and 3 from both sides,
Solve the equation
In solving equations of this type remeber that the line separating the numerator and denominator act as a bracket. The LCM of the denominators 3 and 2 is 6. Multiplying each term of the equation by gives:
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M1 Solve the following equations: 1. x + 3 = 8
17.
2. x − 4 = 6 3. 2x = 8 4. 2x − 7 = 9 5. 5x + 3 = 18 6. 3x − 7 = x − 5 7. 9 − 2x = 3x + 7 8. 4x − 3 = 6x − 9 9. 5x − 8 = 3x + 2 10.2(x + 1) = 9 11.5(x − 3) = 12 12.3(2x − 1) + 4(2x + 5) = 40
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13.7(2 − 3x) = 3(5x − 1) 14.
15.
16.
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2.9.2.5
Simultaneous Linear Equations
Consider the equations
each equation contains the unknown quantities and which satisfy both equations simultaneously. Equations like these are called simultaneous equations.
2.9.2.6
Therefore the solutions are and . To check these values substitute them in equation (2).
Solution of Simultaneous Linear Equations
The method for solving simultaneous equations is illustrated by the following example. Example: i. Solve the equations (2)
(1)
If we multiply equation (1) by 3 and equation (2) by 5 the coefficient of will be same in both equations.
(3) (4) L N O S E S O P R U P G N I N I A R T R O F
To find the value of we substitute for in either of the original equations. Substituting , in equation (1)
We now eliminate by subtracting equation (3) from equation (4) which gives
ii. Solve the equations (2)
(1)
In these equations it is easier to eliminate because the same coefficient of can be obtained in both equations simply by multiplying equation (2) by 2. (3) Adding equations (1) and (3) gives Substituting in equation (1) gives
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M1 Solve the following equations for x and y and check the solutions: 1. 2x − 3y = −8 x + 3y = 14 2. 3x + 5y = 17 4x + 5y = 21 3. 3x + 4y = 26 x +y =7 4. 5x − 7y = 1 2x + 5y = 16 5.
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2.10 Quadratic Equations An equation which can be written in the form is called a quadratic equation. The constants can take any numerical value. The following are all examples of equadratic equations: in which
in which in which A quadratic equation has two solutions (often called the roots of the equation). It is possible for one of the roots to be zero or for the two solutions to be the same.
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2.10.1
Solving Quadratic Equations
A quadratic equation can be solved by factorisation. We make use of the fact that if the product of two factors is zero, then one of those factors must be zero. Thus if then either or . To solve a quadratic equation by this method the expression is written as the product of two factors.
4. Solve Factorising or For both factors,
Example: 1. Solve Writing the equation as
Either
giving or giving
The roots are and , often written as .
5. Solve
Factorising, Either , giving or giving The roots are and . 3. Solve L N O S E S O P R U P G N I N I A R T R O F
Factorising Either or , giving . The roots are and (Note that it is incorrect to say that the solution is . The solution must also be stated).
, giving
Multiply each term by to clear the fraction, Factorising
2. Solve
With this equation both roots are the same and we say that the equation has equal roots. This always happens when the expression forms a perfect square.
and factorising the LHS
Either , giving or giving The roots are and .
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2.10.2
Equations which will not Factorise
Example:
This equation is called the quadratic formula. Note that the whole of the numerator, including , is divided by . The formula is used when factorisation is not possible, although it may be used to solve any quadratic equation.
i. Solve the equation
and
Example:
Comparing with we have . Substituting these values in the formula gives:
and
The square root of a negative quantity has no arithmetic meaning and it is called an imaginary number. The reason is as follows:
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Hence it is not possible to give a meaning to is said, therefore, to having imaginary roots.
i. Solve the equation
(by using a calculator to find ) ii. Solve the equation
2.10.3 Using the Formula to Solve Quadratic Equations
. The equation
or
or
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M1 Solve each of the following quadratic equations using factorisation:
Solve the following equations by using the quadratic formula:
i. x2 − 4 = 0
i. 4x2 − 3x − 2 = 0
ii. x2 − 16 = 0
ii. x2 − x − 1 = 0
iii. 3x2 − 27 = 0
iii. 3x2 + 7x − 5 = 0
iv. 5x2 − 125 = 0
iv. 7x2 + 8x − 2 = 0
v. (x − 7) ( x + 3) = 0
v. 5x2 − 4x − 1 = 0
vi. x(x + 5) = 0
vi. 2x2 − 7x = 3
vii. (3x − 5) (2x + 9) = 0 viii. x2 − 5x + 6 = 0 ix. x2 + 3x − 10 = 0 x. 6x2 − 11x − 35 = 0
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2.11 Logarithms 2.11.1
Why do we use Logs?
2.11.2
What are Logs?
In the discussion of indices it was noted that whenever a number is “raised” to a power then we write that in exponential notation and the meaning of it is that the number appearing in the base is being multiplied by itself the number of times that is indicated by the exponent. The notation used was such that if we write , what we actually mean is 5 multiplied by itself 3 times.
Continuing on the above reasoning, let us take our simple example again, what number raised to the power of 3 gives 1,000? if we invent an unknown variable, call it and try to write out our question in terms of the notation of algebraic powers we have the following sistuation:
Logarithms are mathematical inventions in order to answer a slightly different, question ( notice the word “invention”, logarithms make certain operations easier to handle and that is all they do, so you should think of them as a definition). In order to motivate why logarithms are introduced in the first place, let us invent a scenario. Suppose someone asked you the following question:
The question is what is in the above formula? How do we solve ? We invent an operation called logarithm, abbreviated to Log and we apply this operation to both sides of the above relation.
What number do I have to raise to the power of 3 in order to get 1,000? this might seem pretty simple and obvious. If you multiply 10 x 10 you 100, and if you muliply 100 x 10 you get 1,000. So, you would say that 10 multiplied by itself 3 times or, in our power notation, is equal to 1,000. Now, this is easy to answer by thinking about powers because the above example is simple powers and simple numbers, and one can reason it out relatively quickly. However, things can get more complicated. Suppose now that you were asked “what number do I have to raise 10 to in order to get 735. All of a sudden the answer is not very obvious. What is so different about this question? L N O S E S O P R U P G N I N I A R T R O F
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There is actually nothing different about this question. You still can try doing the same process, but now the number is not that pretty and it’s not exactly obvious how many times you should multiply 10 by itself to get 735. If you multiply it by itself 2 times you get 100, but 3 times gives 1,000 and you have already exceeded 735! how do we “get out” this power that we need. Logarithms are at the most basic level − invented to answer the general question of how does one extract the base or exponent of an algebraic power when one of these is an unknown.
Log 10x = Log 1000 We then define this new logarithm function to be such that when it applies to a number that is “10 raised to some power” then it literally just gives us the power and the base 10 “disappears”. To reiterate this, what we mean is the following: So, what is the value of Log 1000? How do we know what the right hand side of the equation is? How does this help us at all? Well, this will probably seem magic to you at this point, but one option is to find the (Log) button on your calculator and take the Log 1000 and you get 3. So, , and it is what we expected. How does this help us with anything? It seems like we went in a big loop, and we knew the answer to begin with anyway. But, now consider the slightly more complicated question that we had above. What number do I raise 10 to in order to get 735? Let us apply the logarithm process to this situation: If you take the Log of 735 on your calculator you get 2.866. So, 10 raised to the power of 2.866 gives you 735, and the question is answered. Recall that algebraic powers need not be integers, and we have a clear example of a non−integer power.
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2.11.3
Common Lo Logarithms
There are two basic types of logarithms that are important to know. In the previous section, where logarithms were defined, you already saw the difinition of one kind of logarithms, that was the so called Log Base 10. The logarithmic operation that we have introduced serves the maoin purpose of extracting the exponents in an algebraic power. This is true of the operation of “taking the logarithm”.
2.11.3.1 2.11.3.1 Logarithm Logarithm Base 10 The logarithm of base 10 is most often useful when powers of 10 are involved, but not necessarily. It can be used in many other situations. For instance, suppose you were asked the following question: 3 raised to what power gives 16.8? Again, applying our definition of logarithm of base 10 ass defined in the previous section we can answer the question but, in order to do this we need to define some rules of operation for logarithms. You can think if Logarithm Logarithm Base 10 10 as the logarithmic logarithmic operation operation that when when carried carried out on 10 raised to some power ends up giving us the power. The log of base 10 . This is the basic definition of base 10 is written as : . Thus, . logs. abbreviated to “lg”. The following values may be checked using a calculator: lg 15.4 = 1.1875, lg 378.2 = 2.577 and lg 0.0241 = −1.6179. L N O S E S O P R U P G N I N I A R T R O F
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2.11.3.2 2.11.3.2 Natural Natural Logar Logarithm ithms s There is another logarithm that is also useful (and in fact more common in natural processes). Many natural phenonenon are seen to exhibit changes that are either exponentially decaying (radioactive decay for instance) or exponentially increasing (population growth for example). These exponentially changing functions are written as ex, where x represents the rate of the exponential. In such cases where exponential changes are involved we usually use another kind of logarithm called natural logarithm. The natural log can be thought of a logarithm base−e. What this means is that it is a logarithmic operation that when carried out on e raised to some power gives us the power itself. This logarithm is . labelled with Ln (for “natural log”) and its definition is: Ln . Logarithms having a base of e (where ’e’ is a mathematical constant approximately equal to 2.7183) are called hyperbolic or natural logarithms, and loge is often abbreviated to “ln”. The following values may be checked by using a calculator: ln 4.73 = 1.5539, ln 278.4 = 5.6290 and ln 0.7642 = −0.2689
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2.1 2.11.4 1.4
Rul Rules of Log Logarit arithm hms s
There are three rules of logarithms, which apply to any base. Rule 1. 1. To multilply two numbers
Rule 3. 3. To raise a number to a power The following may be checked using a calculator lg = lg 25 = 1.39794
The following may be checked by using a calculator lg10 = 1 Also lg 5 + lg 2 = 0.69897 + 0.301029
Also 2 lg 5 = 2(0.69897) = 1.39794 hence lg = 2 lg 5.
Hence lg (5 x 2) = lg 10 = lg 5 + lg 2. Rule 2. 2. To divide two numbers The following may be checked using a calculator ln = ln 2.5 = 0.91629 Also ln 5 − ln 2 = ln 2.5 = 1.60943 − 0.69314 = 0.91629 Hence ln = ln 5 − ln 2. L N O S E S O P R U P G N I N I A R T R O F
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2.12 Number Systems 2.12.1
Binary
2.12.1.1 Binary to Decimal Conversions
Every number that can be written in decimal can also be written in another system called Binary. Binary as the main number systems used by computer scientists. The binary number system is a base 2 number system which uses only the digits 0 and 1. It is also a place value system which means that each place represents a power of 2, just as the place represents a power of 10 in the decimal system. Powers of 2: Decimal No: 32 16 eg: 0 0
8 1
4 0
2 1
1 0
The number therefore means: 1x8=8 +1 x 2 = 2 +1 X 0.25 = 0.25 So
=
L N O S E S O P R U P G N I N I A R T R O F
. . .
0.5 0
0.25 1
Example: Convert 1001001 to a Decimal Write down the powers of 2, and the number to be converted below them, as follows: 64 1
32 0
16 0
8 1
4 0
2 0
1 1
Then add all the numbers above the 1’s ie. 64 + 8 + 1 = 73
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2.12.1.2 Decimal to Binary Conversions Example: Convert 271 to Binary Write down the powers of two up to the next higher number (256 in this case) than the number to be converted: 256 128 64 32 16 8 4 2 1 Next write in the first digit 1 under the highest number (256). Subtract the 256 from 271. 271 − 256 = 15 256 1
128
64
32
16
8 1
4 1
2 1
1 1
The decimal number is therefore
2.12.1.3 Adding Binary Numbers Add 1100010 to 1000111 Line up the numbers as shown, and add each column starting from the left (as you would when adding decimal numbers). When two 1’s are added, this would normally be 2. But 2 is not allowed in binary, so write 0 and carry 1 to the next column to the left and include it in the addition of the next column. L N O S E S O P R U P G N I N I A R T R O F
1100010 + 1000111 10101001
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2.12.2
Octal
Every number that can be written in decimal can also be written in another system called octal. Like binary, octal is one of the three main number systems used by computer scientists. The octal number system is base 8 number system which uses only the eight digits 0, 1, 2, 3, 4, 5, 6, and 7. It is also a place value system which means that each place represents a power of 8, just as the place represents a power of 10 in the decimal system. Powers of 8; Decimal No: 4096 eg:
512
64 2
8 3
1 7
. .
. 0.125 0.015625
Thus, an octal number such as:
=
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2.12.2.1 Converting Binary to Octal & Octal to Binary To convert a binary number to an octal number, construct a 3−bit binary / octal lookup table like the one below. Starting at the binary decimal point of the binary number, take the first 3 bits and find the corresponding octal value from the table. Repeat with next 3 bits and so on. If less than 3 bits remain, pad them with 0’s until there are 3 bits. Again use the table.
000 001 010 011 100 101 110 111
3−bit binary 0
octal
1 2 3 4 5 6 7 3−bit binary / octal table
Example: Convert 11010010 to octal. 1. Take the 3 most right bits, 010 and find the corresponding octal value in the above lookup table. The octal value is 2. L N O S E S O P R U P G N I N I A R T R O F
2. Take the next 3 bits, 010. The corresponding octal value from the lookup table is 2 again. 3. Now only 2 bits, 11 of the binary number remain. Pad the left hand side with a 0 to get 011. The corresponding octal from the lookup table is 3.
so,
To convert from octal to binary, write down the binary representation of each octal digit. Note that each octal digit should take up 3 bits. Example: Convert to binary 3 = 011 2 = 010 2 = 010 so,
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2.12.3
Hexadecimal
Every number that can be written in decimal can also be written in another system called hexadecimal. Hexadecimal is the last of the three main number systems used by computer scientists. The hexadecimal number system is a base 16 number system which uses the sixteen digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, & F. Here, we need the extra didgits A, B, C, D, E and F to represent the numbers 10, 11, 12, 13, 14, and 15, since there are no digits in the decimal numeral system to do this. Decimal 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Hexadecimal 0 1
Hexadecimal is also a place value system which means that each place represents a power of 16, just as the place represents a power of 10 in the decimal system:
Powers of 16: Decimal No: 4096 256 16 1 eg.: 3 A
. . 0.0625 F .
Thus, a hexadecimal number such as:
= 2
3 4 5 6 7 8 9 A B C D E F
Decimal / Hexadecimal Table
Note: It is much more difficult to convert from decimal to hexadecimal than it is to convert from hexadecimal to decimal. If you are asked in the exam to do the latter, take each answer provided and convert to decimal, until you get the number in the question.
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2.12.3.1 Converting Hexadecimal to Binary & Binary to Hexadecimal To convert a binary number to an hexadecimal number, construct a 4−bit binary/hexadecimal lookup table like the one below. Starting at the binary decimal point of the binary number, take the first 4 bits and find the corresponding hexadecimal value from the table. Repeat with the next 4 bits and so on. If there is less than 4 bits remaining, pad them out to 4 bits.
4−bit Binary
Hexadecimal
0000
0
0001
1
0010
2
0011
3
0100
4
0101
5
0110
6
0111
7
1000
8
1001
9
1010
A
1011
B
1100
C
1101
D
1110
E
1111
F
4−bit Binary / Hexadecimal Table
Example: Convert 11010010 to Hexadecimal 1. Take the 4 most right bits, 0010 and find the corresponding hexadecimal value in the lookup table. The hexadecimal value is 2. 2. The next 4 bits, 1101 and find the corresponding hexadecimal value in the lookup table. The hexadecimal value is D.
To convert from hexadecimal to binary, write down the binary representation of each hexadecimal digit. Note that each hexadecimal digit should take up 4 bits. so,
Example: Convert to binary 2 = 0010 C = 1100 F = 1111 so,
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2.12.4
Binary Coded Decimal
The BCD system is a 4 −bit system representing a decimal character for use with digital display readouts. It can also be used for addressing to make it more convenient for humans to use. BCD number 1001 Decimal Equivalent
0010 9
0011 2
0000 3
0
Thus, a BCD number such as 1001001000110000 is 4 sets of 4−bit binary numbers 9, 2, 3 and 0, which when decoded means decimal 9230.
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2.13 Test Solve the following examples: Binary Convert the following Binary numbers to Decimals 1. 1101.1 2. 1001110.11 3. 100100.1
Hexadecimal Convert the following Binary bits to Hexadecimal code 1. 11100001 2. 101110001111 3. 11111100
Convert the following Decimal numbers to Binary 1. 62 2. 1,204 3. 42.25 4. 51.125
Convert the following Hexadecimal codes to Binary bits 1. 4F 2. 1AC 3. 67 4. 2A8
Add the following Binary numbers 1. 111 and 100 2. 10010 and 1101 3. 10110001 and 11100010
Convert the folloing Hexadecimal codes to Decimal 1. 2D 2. 1AF 3. 21A 4. 1AE
Octal Convert the following Binary numbers to Octal 1. 101010100 2. 011110100000 3. 111101001 Convert the following Octal numbers to Binary 1. 1263 2. 65217 3. 426 4. 5625
Convert the following Decimal numbers to Hexadecimal codes 1. 1632 2. 494 3. 5174 4. 67
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M1 Binary Code Decimal Convert the following Decimal numbers to BCD 1. 94 2. 429 3. 2947 4. 1736 Convert the following BCD numbers to Decimal 1. 10000101 2. 011100001001 3. 001101100100
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3.
Geometry
3.1 Geometrical Constructions 3.1.1 Triangles 3.1.1.1
Pythagoras Theorem
“In a right angled triangle, the area of the square on the hypotenuse is equal to the sum of the areas of the squares on the other two sides”.
Pythagoras Theorem is particularly useful to you when dealing with vectors in aircraft theory. Example: A right angled triangle has smaller sides of length 5 cm and 12 cm. What is the length of the hypotenuse.
c2 c
b
b2
5cm
a L N O S E S O P R U P G N I N I A R T R O F
12cm
a2
By pythagoras c2 = 122 + 52 = 144 + 25 = 169 For the triangle shown: c2 = a2 + b2 a2 = c2 − b2 b2 = c2 − a2
c = = 13cm
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Pythagoras Theorem can be used to find the height of equilateral and isosceles triangles (an equilateral triangle has all its sides of equal length and an isosceles triangle has 2 sides of equal length).
As an example, we could tip the triangle onto its side and take a new height for it.
Example: Find the height of an isosceles triangle which has sides of length 13 cm and base of length 10 cm.
13cm h2 = 132 − 52 = 169 − 25 2 h = 144 h = 12cm
10cm
13cm
13cm
13cm Note now that the base is 13 cm long, whereas before 10 cm long. The height will also be a different length, so there is not one height for one triangle, it all depends on which side you use as your base.
5cm
5cm
10cm
L N O S E S O P R U P G N I N I A R T R O F
Note: The height is always drawn at right angles to the base and goes to the opposite apex. We can draw the height from any side providing it meets the above requirement, i.e it cuts the chosen base at right angles and goes from the base to the opposite apex.
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3.1.1.2
Area of a Triangle
To calculate the area of a triangle, we can use the following formula; Area = x base x height Any side can be chosen as base providing you use the corresponding height.
3.1.1.3
Area of a Right−Angled Triangle
The height of a right−angled triangle is just one of its sides. Example:
Example: Using our first example, we had height = 12 cm and base = 10 cm
Area = x base x height = x 10 x 12 = 60 cm2
Height
In our second triangle, the lengths of all the sides are the same as our first triangle, i.e it is the same triangle and therefore has the same area. We can use this fact to calculate the new height that corresponds to have a base of 13 cm. Area = 60 = x base x height 60 = x 13 x height h = h = 9.23 cm
L N O S E S O P R U P G N I N I A R T R O F
Base Area = x base x height = x 3 x 4 = 6 cm2 Obviously, we could tip the triangle on its side. The 4 cm side now becomes the base and the 3 cm side becomes the height. Note, no difference in size of area.
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M1 Using Pythagoras Theorem find the length of the missing side: 1. In a right angled triangle, triangle, the length of the shorter sides are 8 cm and 15 cm. Calculate the length of the hypotenuse. 2. The hypotenuse of a right angled triangle has has a length of 41 mm. mm. One of the other sides is 9 mm long. Calculate the length of the third side. 3. A right angled triangle has sides of 30 cm and 40 cm. If the third side is the longest what is the length. 4. All the sides of a triangle triangle are 6cm length. What is the vertical height height of the triangle. 5. A triangle has sides which which are 12.5 cm, 30 cm and 32.5 cm in length. Is it a right angled triangle.
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3.2 3.2 Trigo rigono nome metr try y 3.2.1 Trigonometrical Trigonometrical ratios By using pythagoras, you are now able to partially solve right−angled triangles, i.e you can find the third side of a right−angled triangle when given its other 2 sides. This chapter is concerned with establishing the basic trigonometrical concepts which will later enable you to completely solve right−angled triangles, i.e to find all their 6 elements (angles and sides). Similar triangles, are triangles which are the same shape, one is simply an enlargement of the other. Two important properties of similar triangles are: 1. their corresp correspondin onding g angles are equal equal 2. their correspo corresponding nding sides sides are proportional proportional Consider the triangles:
Now consider the following similar triangles, In both cases side ’c’ is the hypotenuse. Taking angle A as the reference: 1. Side ’a’ ’a’ is the the side side opposite opposite 2. Side ’b’ is the adjace adjacent nt Taking angle B as the reference:
L N O S E S O P R U P G N I N I A R T R O F
The above triangles are similar since they are equiangular and the ratios of their corresponding sides are constant, i.e. 1. = 2. 3.
1. Side ’b’ ’b’ is the the side side opposite opposite 2. Side ’a’ ’a’ is the the adjacent adjacent side Since the triangles are similar, the ratios of corresponding sides are constant, i.e, the ratios , and are the same for all similar right−angled triangles.
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M1 In a right−angled triangle the ratio: 1.
is callled the Sine of Sine of the reference angle
sin A = 2.
3.
For the triangle shown: 1. sine sine of of angle angle B 2. cosine cosine of angle angle B 3. tangen tangentt of a angl ngle eB
is called the Cosine of Cosine of the reference angle
cos A =
is called the Tangent of the angle
tan A =
The above are the fundamental trigonometrical ratios for right−angled triangles and must be remembered. A convenient method to help you to remember them is SOHCAHTOA, SOHCAHTOA, where S = sin, C = cos and T = tan.
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1. sin sin B =
2. cos cos B =
3. tan tan B =
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We will now calculate the values for 30 and 60 . Consider the equilateral triangle ABC of sides 2 units.
1. sin 60 =
Line BD bisects ABC and is perpendicular to AC
In triangle ABD,
A = 60 , B = 30 and D = 90
side d = 2 (given) Side b = 1 (half of AC) Side a2 = 22 − 12 a2 = 3 L N O S E S O P R U P G N I N I A R T R O F
Thus in a right angled triangle ABD:
2. cos 60 =
3. tan 60 =
4. sin 30 =
5. cos 30 =
6. tan 30 =
We can now collect all our information and show graphically how the basic trigonometrical ratios change as the angle increases from zero to 90. The functions all give graphs which are important. You should know how to sketch them and know how to use them.
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3.2.2 The Sine Curve The Tangent Curve
The Cosine Curve
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Example: The following example involves the use of trigonometry, or combinations of trigonometry and pythagoras, to solve right−angled triangles. In the right−angled triangle ABC, find angle A and side c. Angle A
c b 5 Since
= tan, this is the ratio we use
tan A =
tan A = tan A = 2.4 L N O S E S O P R U P G N I N I A R T R O F
A = 67 .23 (after using a calculator) Side c by pythagoras, c2 = a2 + b2 c = c = = 13
a
12
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M1 For triangle shown, find the sine, cosine and tangent of angles BAC and ABC.
A
B
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C
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3.3 Coordinates & Graphs 3.3.1 Coordinates An equation involving two variables can be represented by a graph drawn on “coordinate axes”. Coordinate axes (see below) consist of a horizontal line (referred to as the x axis) and a vertical line (referred to as the y axis). The point of intersection of these two lines is called the origin ( denoted by the letter “O“).
Along the x and y axes we can mark off units of measurement (not necessarily the same on both axes). The origin takes the value zero on both axes. The x axis takes positive values to the right of th origin and negative values to the left of the origin. The y axis takes positive values above the origin and negative values below the origin.
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Any point on this diagram can be defined by its coordinates (consisting of two numbers). The first, the x coordinate, is defined as the horizontal distance of the point from the y axis, the second, the y coordinate, is defined as the vertical distance of the point from the x axis. In general, a point is defined by its coordinates which are written in the form (a, b).
Example: The point (3, 2) may be plotted on the coordinate axes as follows
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3.3.2 Graphs An equation involving two variables can be represented, on coordinate axes, by means of a graph. For a given range of values of x, the corresponding y values can be calculated from the equation being considered. The points obtained can then be plotted and joined together to form the graph. Before ploting the points on a graph, the axes must be drawn in a way that takes into account the range of the x−values and the range of the y−values. If graph is used (which is desireable) you should use a scale that involes a sensible number of units per square i.e you should use steps of, for example, 1, 2, 5 or 10 etc. units per square depending on the question. You should avoid using steps along the axes of, for example 7 or 9 units per square as this can complicate the graph unnecessarily. Example: Draw the graph of y = 2x + 1 between x = 0 and x = 5 By taking the x values 0, 1, 2, ........5, we can calculate the corresponding y values, as shown below, by first evaluating the component parts of the equation. x: 2x +1 y:
0 0 1 1
1 2 1 3
2 3 4 4 6 8 1 1 1 5 7 9
5 10 1 11
We then plot the points obtained, each point being defined by its x coordinate and its corresponding y coordinate. The points are then joined together to the graph. L N O S E S O P R U P G N I N I A R T R O F
The value of y therefore depends on the value allocated to x. We therefore call y the dependent variable. Since we can give x any value, we call x the independent variable. It is usual to mark the values of the independent variable along the horizontal axis (x). The dependent variable values are marked off along the vertical axis (y). Equations of the type y = 2x + 1, where the highest powers of the variables, x and y, are the first are called equations of the first degree. All equations of this type give graphs which are straight lines and hence they are often called linear equations. In order to draw graphs of linear equations we need only take two points, however three points are adviseable.
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M1 Draw graphs of the following functions taking values of x between −3 and 4. 1. y = 2x + 5 2. y = 3x − 5 3. y = x2 − 6x + 5 (find the roots) 4. y = 4x3 − 16x2 − 16x + 64 (find the roots)
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3.3.2.1
The Straight Line Graph
A straight line is defined as the shortest distance between two points.
Example:
The equation of a straight line is given by: y = mx + c Where m represents the slope of the line and c is the point where the line crosses the y axis (they intercept). The point where the line crosses the x axis is called the x intercept. m = Gradient of the line c = Intercept on the y axis In this example m = −3 and c = 6 As c = 6, we know that this line cuts the y axis at y = 6 (this can be verified by substituting x = 0 into the equation of the line, as x = 0 along the y axis) Similarly, as y = 0 along the x axis, we can substitute y = 0 into the equation of the line to find where the intersects with the axis (the intercept).
L N O S E S O P R U P G N I N I A R T R O F
Note: in this example m = 2 and c = 0, whenever c = 0 the line will pass through the origin.
we have, when y=0 6 − 3x = 0 3x = 6 x =2 Hence the line cuts the x axis at x = 2. We can now say that the y intercept = 6 and the x intercept = 2.
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M1 Example: A striaght line parallel to the x axis takes the form y = constant. Similarly, a straight line parallel to the y axis takes the form x = constant. These case are illustrated below:
In this example, m = 4 and c = −2. We know, immediately that the intercept is −2 (the value of c). To find the x intercept, we substitute y = 0 into the equation of the line. 0 = −2 + 4 4x = 2 x = 0.5 Hence the x intercept is x = 0.5. L N O S E S O P R U P G N I N I A R T R O F
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3.3.2.2
Derivation of the Equation of a Straight Line Graph
Given the coordinates of two points (x1, y1) and (x2, y2) say, we can calculate the equation of the straight line that passes through these points.
Now (1) and (2) give us equations in two unkowns, m and c, ( simultaneous equations) which we can solve. We have 4 = m + c (1) 10 = 3m + c (2) Subtracting (1) from (2) to eliminate c we obtain 6 = 2m m=3 Substituting this value of m back into (1) we obtain 4=m+c 4=3+c c = 4 − 3 c=1
There are two methods of calculating this equation. Example: The question is; Find the equation of the striaght line that passes through the points (1, 4) and 3, 10).
L N O S E S O P R U P G N I N I A R T R O F
Method 1: The general equation of a straight line is given by y = mx + c and it is necessary to find numerical values for m and c. If the straight line in question passes through the two given points, then each of these points must satisfy the equation of this straight line. That is, we can substitute the coordinates of each point as follows: y = mx + c 4=m+c
(1)
Likewise, substituting (3, 10) we have 10 = 3m + c (2)
If we now substitute these numerical values of m and c into the equation y = mx + c, we obtain the equation of the straight line passing through the points (1, 4) and (3, 10). That is: y = 3x + 1
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M1 Method 2: In general, we can consider any two points (x1, y1) and (x2, y2). The straight line passing through these points can be written as: y − y1 = m (x − x1) where m = (y1 − y2) (m is the gradient of the line) (x2 − x1) Applying this points (1, 4) and (3, 10) we have x1 = 1, y1 = 4, x 2 = 3, y2 = 10 and we hence obtain: m = 10 − 4 = 6 = 3 3 − 1 2 and our line becomes: y − 4 = 3(x − 1) y − 4 = 3x − 3 y = 3x + 1
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3.4 Geometry − Angles 3.4.1 General
3.4.2 Degrees & Radians − Measuring Angles
We can specify an angle by using a point on each ray and the vertex. The angle below may be specified as angle ABC or as angle CBA; you may also see this written as ABC or as CBA how the vertex point is always given in the middle.
We measure the size of angle using degrees. We can also use radians to measure angles. There are 2 radians in 360 . The radius of a circle fits around the circumference 6.26 (or 2 ) times. 1 radian = 57.3 degrees. To convert from degrees to radians, use: x 2 , where is the number of degrees. 360 Example: Here are some examples of angles and their degree measurements. Convert them to radians.
Many different names exist for the same angle. For the angle below, PBW, CBP and WBA are all names for the same angle.
L N O S E S O P R U P G N I N I A R T R O F
PBC,
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3.4.3 Acute Angles
3.4.4 Obtuse Angles
An acute angle is an angle measuring between 0 and 90 .
An obtuse angle is an angle measuring between 90 and 180 .
Example: The following angles are all acute angles
Example: The following angles are all obtuse
3.4.5 Reflex Angles L N O S E S O P R U P G N I N I A R T R O F
A reflex angle is an angle measuring between 180 and 360 degrees.
3.4.6 Right Angles A right angle is an angle measuring 90. Two lines or line segments that meet at right angle are said to be perpendicular. Note that any two right angles are supplementary angles (a right angle is its own angle supplement).
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3.4.7 Complementry Angles
3.4.8 Supplementary Angles
Two angles are called complementary angles if the sum of their degree measurements equal 90 . One of the complementary angles is said to be the complement of the other.
Two angles are called supplementary angles if the sum of their degree measurements equal 180 . One of the supplementry angles is said to be the supplementary of the other.
Example: These two angles are complementary
Note that these two angles can be “pasted” together to form a right angle.
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Example: These two angles are supplementry.
Note that these two angles can be “pasted” together to form a straight line.
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3.4.9 Vertical Angles For any two lines that meet, such as in the diagram below, angle AEB and angle DEC are called vertical angles. Vertical angles have the same degree measurement. Angle BEC and angle AED are also vertical angles.
3.4.11
Alternate Exterior Angles
For any pair of parallel lines 1 and 2, that are both intersected by a third line, such as line 3 in the diagram below, angle A and angle D are called alternate angles. Alternate angles have the same degree measurement. Angle B and angle C are also alternate exterior angles.
3.4.10
Alternate Interior Angles
For any pair of parallel lines 1 and 2, that are both intersected by a third line, such as line 3 in the diagram below, angle A and angle D are called alternate interior angles. Alternate interior angles have the same degree measurement. Angle B and angle C are also alternate interior angles.
L N O S E S O P R U P G N I N I A R T R O F
3.4.12
Corresponding Angles
For any pair of parallel lines 1 and 2, that are both intersected by a third line, such as line 3 in the diagram below, angle A and angle C are called corresponding
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angles. Corresponding angles have the same degree measurement. Angle B and angle D are also corresponding angles.
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3.4.13
Angle Bisector
An angle bisector is a ray that divides an angle into two equal angles. Example: The central ray on the right is the angle bisector of the angle on the left.
The centre ray on the right is the angle bisector of the angle on the left.
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3.5 Circles 3.5.1 Chords & Radii All the “parts” of a circle, such as the radius, the diameter, etc, have a relationship with the circle or another “part” that can always be expressed as a theorem. The two theorems that deal with chords and radii (pulral raduis) are outlined below.
ii. In a circle or in congruent circles, if two chords are the same distance from the centre, then they are congruent. Using these theorems in action is seen in the example below. Find CD.
i. If a radius of a circle is perpendicular to a chord, then the radius bisects the chord. Here’s a graphical representation of this theorem:
OC = radius AB = chord
Given: circle R is congruent to circle S. chord AB = 8 L N O S E S O P R U P G N I N I A R T R O F
RM = SN Solution − By theorem number ii above, segment AB is congruent to segment CD. Therefore, CD equals 8.
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3.5.2 Tangents
3.5.3 Congruent Arcs
The tangent being discussed here is not the trigonometrical ratio. This kind of tangent is a line or line segment that touches the perimeter of a circle at one point only and is perpendicular to the radius that contains the point.
Congruent arcs are acrs that have the same degree measure and are in the same circle or in congruent cirles. Arcs are very important and let us find out a lot about circles. Two theorems involving arcs and their central angles are outlined below.
Example: Find the value of x. Given: segment AB is tangent to circle C at B.
1. For a circle or for congruent circles, if two minor arcs are congruent, then their central angles are congruent. 2. For a circle or for congruent circles, if two central angles are congruent then their arcs are congruent. Example:
x
L N O S E S O P R U P G N I N I A R T R O F
Solution: x is a radius of the circle. Since x contains B, and AB is a tangent segment, x must be perpendicular to AB (the definition of a tangent tells us that). If it is perpendicular, the triangle formed by x, AB and CA is a right triangle. Use Pythagoras Theorem to solve for x.
Arc PQ and arc QR are congruent. Angle POQ and angle QOR are congruent (they are supplemental and since POQ = 90 ). Theorem i. Angle POQ and angle QOR are congruent. Arc PQ and arc QR have to be congruent by theorem ii.
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3.5.4 Inscribed Angles An inscribed angle is an angle with its vertex on a circle and with sides that contain chords of the circle. The figure below shows an inscribed angle.
Example: Find the measure of each arc or angle listed below. Note; A right angled triangle fitted inside a semicircle, the other 2 angles must be 30 and 60 . arc QSR; angle Q and angle R.
The most important theorem dealing with inscribed angles is stated below. The inscribed angle is equal to its intercepted arc. Solution: Arc QSR is 180 because it is twice its inscribed angle( angle QPR, which is 90 ) L N O S E S O P R U P G N I N I A R T R O F
Angle Q is 60 because it is half of its intercepted arc, which is 120. Angle R is 30 by the triangle Sum Theorem which says a triangle has three angles that equal 180 when added together. In the last problem, you noticed that angle P is inscribed in semicircle QPR and angle P = 90 . This leads to our next theorem, which is stated below. Any angle inscribed in a semicircle is a right angle.
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The one last theorem dealing with inscribed angles is a bit more complicated because it deals with quadrilaterals too. If a quadrilateral is inscribed in a circle, then both pairs of opposite angle are supplementary. Example Find the measure of arc GDE.
Solution: By the theorem stated above, angle D and angle F are supplementary. Therefore, angle F equals 95 . The first theorem discussed in this section tells us the arc is twice that of its inscribed angle. With that theorem, arc GDE is 190 .
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3.5.5 Circumference & Arc Length When dealing with circles the circumference, or the distance around a circle. The circumference of a circle equals 2 times times the radius. This is usually represented by the following equation (where C represents circumference and r stands for radius): C = 2 r For example, if a circle has a radius of 3, the circumference of the circle is 6 . Also, you can find the length of any arc when you know its angle and the radius with the following formula. L = length, n = degree measure of arc, r = radius,
L = 2 r
Example: find the length of a 24 arc of a circle with a 5 cm radius.
L N O S E S O P R U P G N I N I A R T R O F
Solution: L = 2 r = 2 5 = the length of the arc is cm, or 2.1 cm
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TABLE OF CONTENTS M1
MATHEMATICS . . . . . . . . . . . . . . . . . . . .
1
1.9.1
Converting Decimals to Fractions . . . . . . . . . . . . . . . . .
20
1.9.2
Converting Fractions to Decimals . . . . . . . . . . . . . . . . .
20
Converting Fractions to Percentages . . . . . . . . . . . . . .
21
1.
ARITHMETIC . . . . . . . . . . . . . . . . . . . . . . .
2
1.9.3 1.9.4
Converting Percentages to Fractions . . . . . . . . . . . . . .
21
1.1
General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
1.9.5
Converting Percentages to Decimals . . . . . . . . . . . . . .
21
1.2
Addition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
1.3
Subtraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
1.9.6 1.9.6.1 1.9.6.2
Converting Decimal to a Percentage . . . . . . . . . . . . . . . Values of a Percentage of a Quantity . . . . . . . . . . . . . . . . Expressing one Quantity as a Percentage . . . . . . . . . . . .
22 22 22
1.4
Multiplication . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4
1.10
Ratio & Proportion . . . . . . . . . . . . . . . . . . . . . . .
24
1.5
Division . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
1.10.1
Ratio . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
24
1.6
Signed Numbers . . . . . . . . . . . . . . . . . . . . . . . . .
8
1.6.1
Adding Signed Numbers . . . . . . . . . . . . . . . . . . . . . . . . .
8
1.6.2
Subtracting Signed Numbers . . . . . . . . . . . . . . . . . . . . .
8
1.6.3
Multiplying Signed Numbers . . . . . . . . . . . . . . . . . . . . . .
8
1.10.2 1.10.2.1 1.10.2.2 1.10.2.3 1.10.2.4
Proportion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Direct Proportion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Inverse Proportion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Proportional Parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
25 25 25 25 26
1.6.4
Dividing Signed Numbers . . . . . . . . . . . . . . . . . . . . . . . .
8
1.11
Power and Roots . . . . . . . . . . . . . . . . . . . . . . . .
28
1.7
Common Fractions . . . . . . . . . . . . . . . . . . . . . .
10
1.12
Indices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
29
1.7.1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10
1.12.1
Base, Index & Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
29
1.7.2
Lowest Terms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10
1.7.3
Comparing the Size of Fractions . . . . . . . . . . . . . . . . . .
12
1.7.4
Adding & Subtracting Fractions . . . . . . . . . . . . . . . . . . .
12
1.7.5
Multiplication of Fractions . . . . . . . . . . . . . . . . . . . . . . . .
14
1.7.6
Division of Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
15
1.8
Decimals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
18
1.12.2 1.12.2.1 1.12.2.2 1.12.2.3 1.12.2.4 1.12.2.5 1.12.2.6
Laws of Indices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Multiplication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Division of Powers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Powers of Powers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Zero Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Negative Indices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Fractional Indices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
30 30 30 30 31 31 31
1.8.1
Adding Decimals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
18
1.13
Tranposition of Formulae . . . . . . . . . . . . . . . .
34
1.8.2
Subtracting Decimals . . . . . . . . . . . . . . . . . . . . . . . . . . . .
19
1.8.3
Multiplying Decimals . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
19
1.14
Areas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
36
1.8.4
Dividing Decimals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
19
1.15
Volumes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
40
1.9
Conversions . . . . . . . . . . . . . . . . . . . . . . . . . . . .
20
1.16
Conversions . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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TABLE OF CONTENTS 1.17
Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
47
2.
ALGEBRA . . . . . . . . . . . . . . . . . . . . . . . . .
48
2.1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
48
2.2
Use of Symbols . . . . . . . . . . . . . . . . . . . . . . . . .
48
2.3
Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
50
2.4
Addition & Substraction of Algebraic Terms
52
2.5
Multiplication & Division Signs . . . . . . . . . . .
52
2.6
Muliplication & Division of Algebraic Quantities . . . . . . . . . . . . . . . . . . . . .
56
2.7
Brackets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
60
2.8
Factorisation . . . . . . . . . . . . . . . . . . . . . . . . . . . .
64
2.8.1
Factorising by Grouping . . . . . . . . . . . . . . . . . . . . . . . . .
64
2.8.2
Factors of Quadratic Expressions . . . . . . . . . . . . . . . . .
66
2.8.3 2.8.3.1 2.8.3.2
Algebraic Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Adding & Subtracting Algebraic Fractions . . . . . . . . . . . . . Multiplication & Division of Algebraic Fractions . . . . . . . .
68 70 70
2.9
Linear Equations . . . . . . . . . . . . . . . . . . . . . . . .
2.9.1
Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.9.2 2.9.2.1 2.9.2.2 2.9.2.3
2.9.2.5 2.9.2.6
Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72 Solving Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 72 Equation Requiring Multiplication & Division . . . . . . . . . . . 73 Equations Requiring Addition & Subtraction . . . . . . . . . . . 73 2.9.2.4 Equations Containing the Unknown Quantity on Both Sides . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74 Simultaneous Linear Equations . . . . . . . . . . . . . . . . . . . . . 76 Solution of Simultaneous Linear Equations . . . . . . . . . . . 76
2.10
Quadratic Equations . . . . . . . . . . . . . . . . . . . . .
2.10.1
Solving Quadratic Equations . . . . . . . . . . . . . . . . . . . . .
79
2.10.2
Equations which will not Factorise . . . . . . . . . . . . . . . .
80
2.10.3
Using the Formula to Solve Quadratic Equations . . .
80
2.11
Logarithms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
82
2.11.1
Why do we use Logs? . . . . . . . . . . . . . . . . . . . . . . . . . . . .
82
2.11.2
What are Logs? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
82
2.11.3 2.11.3.1 2.11.3.2
Common Logarithms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Logarithm Base 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Natural Logarithms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
83 83 83
2.11.4
Rules of Logarithms . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
84
2.12
Number Systems . . . . . . . . . . . . . . . . . . . . . . . .
86
2.12.1 2.12.1.1 2.12.1.2 2.12.1.3
Binary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Binary to Decimal Conversions . . . . . . . . . . . . . . . . . . . . . . Decimal to Binary Conversions . . . . . . . . . . . . . . . . . . . . . . Adding Binary Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . .
86 86 87 87
2.12.2 2.12.2.1
Octal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Converting Binary to Octal & Octal to Binary . . . . . . . . . .
88 89
2.12.3
Hexadecimal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90 2.12.3.1 Converting Hexadecimal to Binary & Binary to Hexadecimal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91
72
2.12.4
Binary Coded Decimal . . . . . . . . . . . . . . . . . . . . . . . . . . . .
92
72
2.13
Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
94
3.
Geometry . . . . . . . . . . . . . . . . . . . . . . . . . .
96
3.1
Geometrical Constructions . . . . . . . . . . . . . . .
96
3.1.1 3.1.1.1 3.1.1.2 3.1.1.3
Triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Pythagoras Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Area of a Triangle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Area of a Right− Angled Triangle . . . . . . . . . . . . . . . . . . . . .
96 96 98 98
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TABLE OF CONTENTS 3.2
Trigonometry . . . . . . . . . . . . . . . . . . . . . . . . . . . .
100
3.2.1
Trigonometrical ratios . . . . . . . . . . . . . . . . . . . . . . . . . . . .
100
3.2.2
The Sine Curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
103
3.3
Coordinates & Graphs . . . . . . . . . . . . . . . . . . .
3.3.1
Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
106
3.3.2 3.3.2.1 3.3.2.2
Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The Straight Line Graph . . . . . . . . . . . . . . . . . . . . . . . . . . . . Derivation of the Equation of a Straight Line Graph . . . .
107 110 112
3.4
Geometry − Angles . . . . . . . . . . . . . . . . . . . . . .
114
3.4.1
General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
114
3.4.2
Degrees & Radians − Measuring Angles . . . . . . . . . . .
114
3.4.3
Acute Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
115
3.4.4
Obtuse Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
115
3.4.5
Reflex Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
115
3.4.6
Right Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
115
3.4.7
Complementry Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . .
116
3.4.8
Supplementary Angles . . . . . . . . . . . . . . . . . . . . . . . . . . .
116
3.4.9
Vertical Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
117
3.4.10
Alternate Interior Angles . . . . . . . . . . . . . . . . . . . . . . . . .
117
3.4.11
Alternate Exterior Angles . . . . . . . . . . . . . . . . . . . . . . . . .
117
3.4.12
Corresponding Angles . . . . . . . . . . . . . . . . . . . . . . . . . . .
117
3.4.13
Angle Bisector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
119
3.5
Circles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.5.1
Chords & Radii . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
121
3.5.2
Tangents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
122
3.5.3
Congruent Arcs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
122
3.5.4
Inscribed Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
123
3.5.5
Circumference & Arc Length . . . . . . . . . . . . . . . . . . . . . .
125
106
121
