3.Easa Part 66 Mod.1 3

Integrated Training System Designed in association with the club66pro.com question practice aid

TTS Integrated Training System Module 1 Licence Category B 1 and 82 Mathematics 1.3 Geometry

~

U se and/or disclosure is govemed by th e s1a1emen1

on page 2 of this chapter

Module 1.3 Geometry

3-1 TTS Integrated Training System ©Copyright 2011

.


Copyright Notice © Copyright. All worldwide rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form by any other means whatsoever: i.e. photocopy, electronic, mechanical recording or otherwise without the prior written permission of Total Training Support Limited.

Knowledge Levels- Category A, 81, 82 and C Aircraft Maintenance Licence Basic knowledge tor categories A, B1 and B2 are indicated by the allocation of knowledge levels indicators (1, 2 or 3) against each applicable subject. Category C applicants must meet either the category B1 or the category B2 basic knowledge levels. The knowledge level indicators are defined as follows:

---..,

LEVEL 1 A familiarisation with the principal elements of the subject. Objectives: The applicant should be familiar with the basic elements of the subject. The applicant should be able to give a simple description of the whole subject, using common words and examples. The applicant should be able to use typical terms.

LEVEL 2 A general knowledge of the theoretical and practical aspects of the subject. An ability to apply that knowledge. Objectives: The applicant should be able to understand the theoretical fundamentals of the subject. The applicant should be able to give a general description of the subject using, as appropriate, typical examples. The applicant should be able to use mathematical formulae in conjunction with physical laws describing the subject. The applicant should be able to read and understand sketches, drawings and schematics describing the subject. The applicant should be able to apply his knowledge in a practical manner using detailed procedures.

LEVEL 3 A detailed knowledge of the theoretical and practical aspects of the subject. A capacity to combine and apply the separate elements of knowledge in a logical and comprehensive manner. Objectives: The applicant should know the theory of the subject and interrelationships with other subjects.

The applicant should be able to give a detailed description of the subject using theoretical fundamentals and specific examples. The applicant should understand and be able to use mathematical formulae related to the subject. The applicant should be able to read, understand and prepare sketches, simple drawings and schematics describing the subject. The applicant should be able to apply his knowledge in a practical manner using manufacturer's instructions. The applicant should be able to interpret results from various sources and measurements and apply corrective action where appropriate .

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Module 1 .3 Geometry

Use and/or disclosure is governed by th e statement on page 2 of this chapter

,....--,

- . ~') ~:.r/ '~;/


Table of Contents

,-.

~

-~

.-----..

r--'

Module 1.3 Geometry

5

Trigonometry Trigonometrical Relationships The Sine Curve The Cosine Curve The Tan Curve Other Trigonometric Functions To Find the Length of an Unknown Side Coordinates and Graphs The x and y Axis Graphical Representations of an Equation The Straight Line Derivation of the Equation y = mx + c Cartesian and Polar Coordinates Cartesian Coordinates Polar Coordinates Converting

Use and/or disclosure is governed by the statement on page 2 of this chapter

5

5 10 10 11 12 13 21

21 27 39

45 51

51 51 52

Module 1.3 Geometry

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Module 1.3 Enabling Objectives and Certification Statement Certification Statement These Study Notes comply with the syllabus of EASA Regulation 2042/2003 Annex Ill (Part-66) . t ed K nowe I d~e Leve Is as spec1T1ed beow: I A ppen d.1x I, an d th e assoc1a EASA66 Level Objective Reference 81 82 Geometry 1.3 (a) 1 1 Simple geometrical constructions (b) 2 2 Graphical representations; nature and uses of graphs, graphs of equations/functions 2 2 (c) Simple trigonometry; trigonometrical relationships, use of tables and rectangular and polar coordinates


Module 1.3 Geometry

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Module 1.3 Geometry Trigonometry Trigonometrical Relationships 1.

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).

2.

Similar triangles are triangles which are the same shape, one is simply an enlargement of the other. Two important properties of similar triangles are: a)

their corresponding angles are equal.

b)

their corresponding sides are proportional. Consider the triangles: (2)

(1)

B

3

c 8

4 '"'

3.

The above triangles are similar since they are equiangular and the ratios of their corresponding sides are constant, i.e.

3

EF

5

DE

4 5

OF

-

b)

-=-= -

c)

-

,.---.,

;---.

,...--.

BC

a)

~


AB AC AB

BC

AC

=- =-

3 4

- --

DE

EF

-

DF

6

3

10

5

8 10

= -5

= -= -

=-

6

4

3 4

= - -- -

8

Module 1.3 Geometry

3-5 TIS Integrated Training System ©Copyright 2011


8

8

a

c

A

b

4.

b

Now consider the following similar triangles: In both cases side 'c' is the hypotenuse. Taking angle A as the reference:

Taking angle Bas the reference:

a) b)

a) b)

Side 'a' is the side opposite Side 'b' is the adjacent side

Side 'b' is the side opposite Side 'a' is the adjacent side

Since the triangles are similar, the ratios of corresponding sides are constant, i.e., the ratios a , b and a are the same for all similar right-angled triangles. c c b 5.

In a right-angled triangle the ratio: a)

side opposite the angle

sin A =

b)

opposite a = hypotenuse c

side adjacent to the angle hypotenuse cos A=

c)

is called the SINE of the reference angle

hypotenuse

adjacent hypotenuse

side opposite the angle side adjacent that angle

=

is called the COSINE of the reference angle b c

is called the TANGENT of the angle.

tan A = opposite = a adjacent b 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' or 'SoHCAHToA' where S=sin, C=cos and T=tan


Module 1.3 Geometry



Example: For the triangle shown find: a)

sine of angle B

b)

cosine of angle B

c)

tangent of angle B

4

A

6.

=

a)

sin B

b)

cos B

=

c)

tan B

=

3

opp hyp

=

adj hyp

=-

opp adj

=

5

4

5

=

c

3

3 4

0.6

=

0 .8

=

0.75

We will now investigate how the values of sin, cos and tan vary with the magnitude of the angle. a)

When angle A is very small: B

a A

(1) sin A

.

=

opp hyp

=

b

a c

.

C

and is very small. When angle A is zero, sin A

(2) cos A

=

adj hyp

=

b

c

and b::: c. When angle A is zero, cos A

Use and/or disclosure is gov<>rned by the statement

on page 2 of this Chapter

=0

Module 1.3 Geometry

=1

3-7 TIS Integrated Training System © Copyright 2011


(3)

tan A

=

opp adj

=

a

and is very small.

b

When angle A is zero, tan A

=

0

b) When angle A is large:

B

a

A

(1) sin A

=

(2) cos A

=

(3) tan A

=

b

opp =a and a hyp c

c

c.

adj =E_ and is small. hyp c When angle A

=

goo, cos A= 0.

opp =~ and is very large. adj b When angle A

= goo,

tan A

==

We can summarise the above: ANGLE sin

0

1

cos

1

0

tan

0

CXl

Note: The maximum value of sin and cos is 1 , but the maximum value of tan is infinity (= ).


Module 1.3 Geometry

Use and/or disclosure is governed by the statement on page 2 o f this ch apter

•'

~-

,--..

r--<

Integrated Training System

,..--...

Designed in association with the club66pro.com question practice aid

r--

8. ,.---.,

We have seen that trigonometrical ratios vary as the angle varies and have calculated values for 0° and 90°. We will now calculate the values for 30° and 60°. Consider the equilateral triangle ABC of sides 2 units.

,--..

B

Line BD bisects ABC and is perpendicular to AC 2

In triangle ABO,

A= 60°,

B = 30° and

B

= 90°

D

sided

= 2 (given)

side b

= 1 (half of AC)

side a

= ~2 2

a

f

(Pythagoras)

= .J3

Thus, in right-angled triangle ABO:

,.__

a)

sin 60° =

b)

cos 60° =

c)

tan 60°

d)

sin 30° =

=

e)

cos 30°

=

f)

tan 30°

=

Use and/or disclosure is governed b y the stat ement


opp hyp adj hyp opp adj opp hyp adj hyp opp adj

= = =

.J3 = 2 1

1.7321 2

=

0.8660

= 0.5000

2

.J3 = 1

1.7321

1

= - = 0.5000 2

= =

J3 = 2 1

.J3 =

1.7321 2

= 0.8660

0.5774

Module 1.3 Geometry



-

Designed in association with the club66pro.com question practice aid

9.

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 all and know how to use them.

The Sine Curve xo 0

sin

X

0.5 0

0

30

60

90

120

150

180

210

240

270

300

330

360

0.00

0.50

0.86

1.00

0.86

0.50

0.00

-0.50

-0.86

-1.00

-0.86

-0.50

0.00

v

v

3p sp gp

"1\

11~0

1 0 1 0

-0 .5

"

The Cosine Curve 0

cos X

0

....,.

v

~

0

0

30

60

90

120

150

180

210

240

270

300

330

360

1.00

0.86

0.50

0.00

-0.50

-0.87

-1.00

-0.87

-0.50

0.00

0.50

0.87

1.00

........ 0.5

2 0 2 0 3 0 3

.......

-1

xo

This is the curve drawn when you put all the figures on the graph from the table above. As you can see, this curve is in a wave form. This wave can continue past 360° and go into the negatives .

~

""'"

' 1\

3)

6)

gf'\

-0 .5 -1

3-10 TIS Integrated Training System © Copyright 201 1

I

0 1 0 1 0 2 0

" '

2~0

/

~

If you look at this curve you can see it is actually the same as the sine curve except it is a different section (i.e. this peaks at 0°Where the sine curve peaks at 90').

3 0 3 0 3E 0

v

../

Module 1.3 Geometry


~


The Tan Curve xo tan

X0

0

30

60

0.00

0.58

1.73

90 00

120

150

180

210

240

-1 .73

-0 .58

0 .00

0 .58

1.73

270 00

300

330

360

-1.73

-0.58

0.00

,..... ., 10

t

I

8

6

4

r---

-v

2

,........_

0

-2

,--.

u

'"'

-4

J

1

r

;;>

~ •u

fU

-v

oU

L

J

In

;u

L'

•u

L•

I

-6 -8 -10

I

r~ ou

iU

The tan curve is very different from the others. It is a non-continuous which breaks as the value at the breaking point (when x=90° or x=270°) is infinity. Again this curve can be continued with the ..JI o section from x=90° to x=270° repeated.

~

..J, ru

I !

l

r""""\

From the curves we can see there is always more than one possible value for any 1 number you are working out the inverse of ( sin- 0.5 = 30°or 150°). The problem is that your calculator only gives you one of the values (the one below 90~. You must remember the curves to find the position of the second angle.

10.

You can , of course, use a graph to find the sin and cos of angles between 0 and 90°. For tan, this is only practical (because of length of axis) up to about 45°. You should note from the curves of y = sin andy = cos that there is a definite relationship between sin and cos, e.g. : a)

sin 30° = cos 60° = 0.5000

b)

sin 45° = cos 45° = 0.7071

c)

sin 60° = cos 30° = 0.8660

d)

sin 80° = cos 10° and so on.





Other Trigonometric Functions Although less often used, other trigonometrical terms can be derived from the basic terms sin and cos. These terms are called cot (cotangent), sec (secant) and cosec (cosecant). They are determined as follows: tan

sin

= cos cos sin

cot =

Reciprocal relations: sin

1

= cosec

cosec

1

= sin 1

cos

= sec

sec

= cos

tan

= cot

cot

= tan

1

1

1

Square relations (also known as the Fundamental Identities): sin 2 + cos2 sec2

-

tan 2

cosec2 - coF

=1 =1 =1

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Module 1.3 Geometry

Use and/or disclosure is governed by lhe slatemenl on page 2 of this chapter


To Find the Length of an Unknown Side 1. r----.

So far we have evaluated the sine, cosine and tangent of angles, given the 3 sides of a right-angled triangle. In the following text it is shown how to solve completely a rightangled triangle, given any side and 2 angles.

,..--.. ;-..

2.

B

,.-...--

,........_ r--

a

,..-...

c

A

From the triangle shown: r-,...--..

a)

sin

=

opp hyp

sin

=

a c

_,......., ,--.. r-"

b)

cos

=

cos

=

adj hyp b c

,.....-.

a

=c

tan

=

opp adj

tan

=

a b

,.-...,

sin

b

=c

cos

,...--..

--------

c)

~

,........._ ,..--....._

,..-...

a ;-..

=b

d)

By Pythagoras:

c2

= a2

+ b2

tan

~

,...--..

Use and/o r disclosure is gove rne d by t he stateme nt


Module 1.3 Geometry

3- 13 TTS Integrat ed Training Sy stem

© Copyright 2011


3.

The following examples involve the use of trigonometry, or combinations of trigonometry and Pythagoras, to solve right-angled triangles.

a)

In the right-angled triangle ABC, find angle A and side c

B

A 5

(1)

To find angle A. Note: In terms of angle a, we are given the side opposite and the side adjacent. Since opp = adj

tan, this is the ratio we use.

tan A

=

opp adj

tan A

=

12 5

tan A = 2.4 A

(2)

=

67° 23' (after using a calculator or tables)

To find side c

Note: If we use trig. to find side 'c', it necessitates our using angle A which we have just found. If we have made an error in calculating angle A, this would also result in an error in side 'c' . By using Pythagoras, we use only given information and thus the possibility of 'carrying' an error is eliminated. ·


Module 1.3 Geometry



c2

Use and/or disclosure is govemed by the statement on page 2 of this chapter

= a2

+ b2

c

~a2 + b2

c

~12 2 + 52

c

~144 + 25

c

~169

c

13

Module 1.3 Geometry



Intentionally Blank


Module 1.3 Geometry



Worksheet

1.

For the triangle shown, find the sine, cosine and tangent of

A and

C.

A

8

r-...

c

B

,........_

6 -----,.---.,

, ........ ,...-.. ~

~

r-'\

Use and/or disclosure is g ov e rned by the stat eme nt


Module 1.3 Geometry



Intentionally Blank

3-18 TTS Integrated Training System ©Copyright 201 1

Module 1.3 Geometry



Answers

1.

sin cos

tan

= 0.6 A = 0.8 A = 0.75

A

C

= 0.8

cos

c

0.6

tan

C

1.33

sin

/\

Use and/or disclosure is governed by the statement

on page 2 ol this chapter




Intentionally Blank


Module 1.3 Geometry

Use and/or disclosure is governed by the sta tement



Coordinates and Graphs The x and y Axis An equation involving two variables can be represented by a graph drawn on 'Coordinates Axes'. Coordinate axes (illustrated below) consist of a horizontal line (usually referred to as the x axis) and a vertical line (usually referred to as they axis). The point of intersection of these two lines is called the origin (usually denoted by the letter '0'). y axis

0

x axis

Along the x andy 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 the 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. 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:y

2 ------ ----.., (3,2) l

0


Module 1.3 Geometry



Example: Consider the following diagram y .A

.D -fi

-5

-4

.?,

-2

,F.

The points A, B, C, D, E and F above are defined by their coordinates as follows: A B

c

(1' 4) (3, 2) (2, 1)

D E

F

(-4, 1) (-5, -3) (3, -2)

.~


Module 1.3 Geometry

Use and/or disclosure is

governed by the statement on page 2 of this chapter


,----.,

Worksheet

1.

Plot the following points on coordinate axes.

i)

(2, 3)

ii)

(1 ,4)

iii)

(5, 0)

iv)

(0, 2)

v)

(3,-1)

vi)

(-2, 4)

vii)

(-1, -3)

viii)

(0,-4)

ix)

(-5, 0)

x)

(-4, 1)

xi)

(-3, -1)

xii)

(3, -3)

Use and/or disclosure is governed by the statement on page 2 ol this chapter




Intentionally Blank

,-

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Module 1.3 Geometry

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on page 2 olthis Chapter


Answers 1.

y

X

)((1. 4)

(- 2, 4)

x{2. 3)

,.(-4. 1)

-5

-4

.., -..)

)(

(-3, _.,)

x(_~ . -3)





Intentionally Blank

r--, 3-26 TTS Integrated Training System ©Copyright 2011

Module 1.3 Geometry

Use and/or disclosure is governed by the statement on page 2 of this ctlapter


Graphical Representations of an Equation 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 plotting 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 paper is used (which is desirable) you should use a scale that involves 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.


Modu le 1.3 Geometry

3-27 TTS Integrated T raining System © Copyright 2011


Example Draw the graph of y

=2x + 1 for 0 :::;; x :::;; 5.

By taking x values of 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 2 1

1 1

3

2 4

3 6

1 5

1 7

4

5

8 1

10 1 11

9

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 form the graph. In this example the points to be plotted are (0, 1), (1, 3), (2, 5), (3, 7), (4, 9), (5, 11 ). G raph of y = 2x + 1

y=2x +l

y

11

10 9

8 7

6 5

4 3 2

1 0


1

2

3

4


5

X

Use and/or disclosure is governed by the statement on page 2 ol this chapter


Example:

Draw a graph of y

=x2 - ax + 12 for 0 ~ X ~ 6

We again take x values covering the given range , and calculate the corresponding from the given equation. x: 0 x2 0 -8x 0 +12 +12 y: 12

1 1 -8 12

5

2 4 -16 +12 0

3 9 -24 +12 -3

4 16 -32 +12 -4

5

y values

6 36 -48 +12 0

25 -40 +12 -3

We now plot the points obtained and join them together to form the graph. In this example the points to be plotted are (0, 12), (1, 5), (2, 0), (3, -3), (4, -4), (5, -3), (6, 0). y

11 10 y = x 2 - Sx + 12 6 4 2

4

5

-2 -4 .(i

N. B. For a more detailed graph we could, of course, include more points. e.g. by taking x values of

0, V2, 1, 1%, 2, 2%, ... ...... .5%, 6

and calculating the corresponding y values, we could plot nearly twice as many points as we did in the above example

Use and/or disclosure is governed by the sta tarrw:mt




©Copyright 2011

'


Example: Draw a graph of y

=x 2 + 1 for -3 ~ x ~ + 3

Again , taking x values covering the given range, we first calculate the corresponding y values from the given equation.

x:

0 0 +1

1 1

2

4 +1

-1 1 +1

+1

+1

5

2

1

2

5

-3

-2

X

9

+1

+1 10

y:

3 9

4

+1 10

We now plot the points obtained and join them together to form the graph . Graph of y

= x2 + l

y y

-3

3-30 ITS Integrated Training System ©Copyright 2011

-2

-1

0

= x2 +

1

Module 1.3 Geometry

2

1

3

X

Use andfor disclosure is govem ed by the statement on page 2 of this chapter

'


Worksheet 1. Draw graphs of the following functions for 0 ~

y = 2x + 5 y = 5x + 1 y = 3x- 5

i) ii) iii) iv) v) vi)

2.

x~ 5

y = x2 - 6x + 5 2

y = x - 7x + 12 y = 3x2 - 21 x + 30

Draw graphs of the following functions for -3 i) ii) iii) iv)

y

~

x~3

= 2x22 + 7

y = 3x - 12 3 y =x - 7 y = 4x3 - 16x2 1

-

16x + 64

v) y = -

x+5

Use and/or disclosure is governed by tha statemenl on page 2 ol this chapter

Module 1.3 Geometry



Intentionally Blank


© Copyright 2011

Module 1.3 Geometry

Use and/or disclosure is governed by tho otatomont on page 2 of this chapter

-

Use andlor disclosure is govorned by tho ctatomcnt


Module 1.3 Geometry



1ii)

y =-~X - 5

EE;-· - _ ,-

!__;.;I]

::!o:------:2 ___

)'

10

y = 3x -5

8

4

2 --~~----~--~~------~----~------~-------~

0

X

-2

y=x

2

-

6x + 5

0 5

1

-3

0

y

5 4

J

y

=x 2 -

6x + 5

_5

2 1 I)

X

_,

-2 -3

-5


Module 1.3 Geometry


governed by lhe Slalemenl on page 2 of lhis chapler

r----,

r----,


r-

Designed in association with the club66pro.com question practice aid r-

v)

r--'\

y = x2 - 7x + 12

"""'"'

I;:

0 12

1 6

2 2

4 0

5 2

y = x2 -7x

+ 12

3 0

y 12

10 ,--

8 6 4

,---

2 0

vi) r-

6 y = 3xL 21x + 30

I~:

0 30

1 12

2 0

3 -{)

X

4 --6

5 0

y

y = 3x 2 - 2lx

+ 30

X

Use and/or disclosure is governed by the statement on page 2 or this chapter

Module 1.3 Geometry



2.

i)

y

I;;

= 2x2 + 7

- 3 -2 25 15

-1 9

0

1

2

7

9

15

y

30

y

= 2x2 + 7

25

5 -3

ii)

-2

y = 3x2 - 12

0

-1

x: y:

2

1

-3

15

-2 0

-1

0

- 9 -12

3

1 -9

X

2

3 15

0

y

20

y =

3xL 12

15 10 5 0 -5

X

-20


Module 1.3 Geometry

Use and/or disclosure is governed by the statement on page 2 of this chapte r


y

3

=X

-7 y

y

~(I

=X

3

-7

l:'i )()

5 L...

-4

iv)

-1

-2

-1

(I

y = 4x3- 16x2- l 6x + 64

X

---=-~..,.;~~--;.~-~7,1~-::64:70_;;-;~;-~~~ -2~~ 3

1-1.:.:.;::....:

y

40

y = 4xL 16xL 16x + 64

20 0 - 20

X

-40 --
-80

-100 -120 -140

- 160

-~



Module 1.3 Geometry



~·)

y= X-<--

5

.l(;

-3

-2

y:

\A.

;..\

-1 y;

0 1/s

2 l/4)

l f:

- ~~

':!

0.1

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Module 1.3 Geometry



The Straight Line A straight line is defined as the shortest distance between two points. The equation of a straight line is given by

y

=mx + c

where m represents the slope of the line and cis the point where the line crosses they-axis (i.e. they intercept). The point where the line crosses the x-axis is called the x intercept. Example: Graph of y = 2x y

y

= 2x

4 ,-

2

In this example, m =2 and c

3

4

X

=0

Note that whenever c = 0, the line will pass through the origin. Example: Graph of y = 6- 3x .

-.l 6

4 y

= 6 -3x

2 0

In this example, m = -3 and c

1

2

3

4

X

=6

As c = 6, we know that this line cuts the y axis at y = 6 (this can be verified by substituting x into the equation of the line, as x = 0 along they axis).

=0

Similarly, as y = 0 along the x axis, we can substitute y =0 into the equation of the line to find where the line intersects with the x axis (i.e. the x intercept). We have, when y

=0 6 - 3x = 0 3x= 6

X=2

Use andfor disclosure is govemed by the statement

on page z of this chapter




Hence the line cuts the x axis at x = 2 We can now say that they intercept= 6 and the x intercept= 2

Example: Graph or y "'" - 2 + 4x

X

In this example, m = 4 and c = -2 We know, immediately, that they intercept is -2 (i.e. the value of c) To find the x intercept, we substitute y = 0 into the equation of the line. i.e.

0 = -2 + 4x 4x= +2 X=

Y2

Hence the x intercept is x = Y2

Special Cases A straight 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 cases are illustrated below:

Straight line parallel to the x axis y

_ _ _....::5::....t--- - - - - y 0


Straight line parallel to they axis y

=

x =3

5 X

0

3

Module 1.3 Geometry

X



Worksheet For each of the following equations identify the gradient and the y-intercept.

y = 4x + 5

i) ii) iii)

y= 9x y=8

iv)

Y=-

v)

y=--

vi)

3y

= 9x + 6

vii)

8y

= x- 8

3x

5

7

6 5x 13

Use and/or disclosure is g o ve rned by the sta tement on page 2 of this chapter

4




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lntAgrated Training System © Copyright 2011

Module 1.3 Geometry



Answers m =gradient, c = y-intercept i)

m = 4, c = 5

ii)

m = 9, c = 0

iii)

m

iv)

m=

v)

ms c6 - -13' - 13

vi)

m = 3, c = 2

vii)

m-J.. - 8' c--1 -


= 0, c = 8 _2._

7 '

c

= ..1.5

Module 1.3 Geometry



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Module 1.3 Geometry

Use and/or disclosure is governed

by the statement



Derivation of the Equation y = mx + c Given the coordinates of two points, (x 1, y 1) and (x2 , y2 ) say, we can calculate the equation of the straight line that passes through these points. y

0

X

Two methods of calculating this equation are illustrated below: ,--...._

Example: The question is: Find the equation of the straight line that passes through the points (1, 4) and (3, 10). 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

substituting (1, 4) we have 4=m

(1)

+C

likewise, substituting (3, 10) we have

10 =3m+ c

(2)

Now (1) and (2) give us two equations in two unknowns, m and c, (i.e. simultaneous equations) which we can solve.


Module 1.3 Geometry

3-45 TIS Integrated T raining System © Copyright 2011

Integrated .Training System

Designed in association with the club66pro.comquestion practice aid We have

4=m+c 10 =3m+ c

(1) (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 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

-

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

where m =

(m is the gradient of the line)

Applying this to the points (1, 4) and (3, 10) we have X1 = 1; Y1 = 4; x2 = 3; Y2 = 10; and we hence obtain:

m=

10 4 3

1

=

6 2

= 3

and our line becomes y -4 y -4 y

.....__

= 3(x- 1)

--..

= 3x- 3

--..

=3x + 1

N.B. In this example, the point (1, 4) corresponded to (x1, y1) and the point (3, 10) corresponded to (x2, Y2). If we had worked through this example with the point (3, 10) corresponding to (x1, y1) , and the point (1, 4) corresponding to (x2, y 2), the answer would have been exactly the same . .


Module 1.3 Geometry


----

Integrated Training System Desigrred in association with the club66pro.com question practice aid

Worksheet

1. Calculate the equation of the straight line that passes through the following points. i. li

(1 , 3) and (3, 7) (0, 2) and (5, 22)

li

(1, -5) and (-1, -9)

I.

2. Calculate the equation of the straight line that passes through the following points: i. ii. iii. iv. v. vi. vii. viii. ix. X.

(1 , 7) and (3, 11) (0, 0) and (1, 6) (3, -2) and (2, 1) (0, 0) and (-2, 8) (6, 1) and (4, -1) (0, -3) and (-2, 1) (2, 6) and (7, 6) (-5, -47) and (-2, -26) (3, 1) and (3, -2) (1 , 11A) and (2, 2%)


gove rned by lhe stalemenl on page 2

ol l~is ~apler

Module 1.3 Geometry

3-47 TTS Integrated Training System ·· © Copyright 2011


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Answers

1. i. y = 2x + 1 11. y = 4x + 2 iii. y = 2x- 7

,.-_,

2.

IV.

y = 2x + 5 y = 6x y = -7x + 2 y =- 4X

V.

y

1. 11.

iii.

IX.

5 y = -2x- 3 y= 6 y = 7x - 12 x =3 (y

X.

y= 3~-

vi. vii. viii.

=X-

Use and/or disclosure is gove rned by the stat eme nt on page 2 of this chapter

= mx + c does not work with lines of infinite gradient)

X

Module 1.3 Geometry



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Module 1.3 Geometry




Cartesian and Polar Coordinates To pinpoint where you are on a map or graph there are two main systems:

Cartesian Coordinates Using Cartesian Coordinates you mark a point by how far along and how far up it is (x and y coordinates):

x

(x,y)

---------f'

I I

IY

I I

Polar Coordinates Using Polar Coordinates you mark a point by how far away, and what angle it is (rand coordinates):

e

(r,e)


Module 1.3 Geometry



Converting To convert from one to the other, you need to solve the triangle:

To Convert from Cartesian to Polar If you have a point in Cartesian Coordinates (x, y) and need it in Polar Coordinates (r, 8), you need to solve a triangle where you know two sides. Example: What is (12, 5) in Polar Coordinates?

~~

·.

2£] e -~~

,~/~ \

L

12

1-

-I

Use Pythagoras Theorem to find the long side (the hypotenuse): 2 r = 122 +5 2 2

r = -../ (12 + 5 ) r = -../ (144 + 25)

= -../ (169) = 13

Use the Tangent Function to find the angle: tan 9=

212

e = tan-1 ~ 12

-

= 22.6°

So, to convert from Cartesian Coordinates (x, y) to Polar Coordinates (r, 8):

r

= -../ (x 2 +f)

6=tan-1 Y.... X


Module 1.3 Geometry



To Convert from Polar to Cartesian If you have a point in Polar Coordinates (r, 8), and need it in Cartesian Coordinates (x, y) you need to solve a triangle where you know the long side and the angle: Example: What is (13, 231 in Cartesian Coordinates?

X

1-

.I

Use the Cosine Function for x:

-

X

cos (231 = 13 Rearranging and solving: X

= 13 X

COS

(231 = 13 X 0.921 = 11.98

Use the Sine Function for y: sin (231

= L13

Rearranging and solving: y

= 13 x sin (231 = 13 x 0.391 = 5.08

So, to convert from Polar Coordinates (r, 8) to Cartesian Coordinates (x, y) :

x y


=r x cos( 8) =r x sin( 8)

Module 1.3 Geometry



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Module 1.3 Geometry



Worksheet 1.

Convert the following cartesian coordinates into polar coordinates: (a) (b) (c)

2.

(3, 4) (10, 10)

(1 0, 0)

Convert the following polar coordinates into cartesian coordinates

r---.

,---.. ~

,........._

(a) (b) (c)

13cm,67.4° 50 m, 60° --Js ft, n/2 radians

~

,.-..

,........... ,~

,..---..,

Use and/or disclosure is governed by the statement on page 2 of thi s chapter

Module 1.3 Geometry



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Module 1.3 Geometry




Answers

1.

I r--

(a) (b) (c)

5, 53° 14.14, 45° 10, 0°

(a) (b) (c)

x

2.

= 5 em, y = 12 em

x = 25 m, y = 43.3 m x =2m. y =2m



Module 1.3 Geometry



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Module 1.3 Geometry



3.Easa Part 66 Mod.1 3

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