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TRIGONOMETRY
INTRODUCTION Trigonometry is the branch of Mathematics which deals with the measurement of angles and sides of a triangle. The word Trigonometry is derived from three Greek roots : ‘trio’ meaning ‘thrice or Three’,’gonia’ meaning an angle and ‘metron’ meaning measure. Infact, Trigonometry is the study of relationship between the sides and the angles of a triangle. Trigonometry has its application in astronomy, geography, surveying, engineering and navigation etc. In the past, astronomers used it to find out the distance of stars and plants from the earth. Even now, the advanced technology used in Engineering are based on trigonometric concepts. In this chapter, we will define trigonometric ratios of angles in terms of ratios of sides of a right triangle. We will also define trigonometric ratios of angles of 00, 300, 450, 600, and 900. We shall also establish some identities involving these ratios.
HISTORICAL FACTS Indian Mathematician have established keen interest in the study of Trigonometry since ages. They are known for their innovation in the use of size instead the use of choid. The most outstanding astronomer has been Aryabhatta. Aryabhatta was born in 476 A.D. in Kerala. He studied in the university of Nalanda. In mathematics, Aryabhatta’s contribution are very valuable. He was the first mathematician to prepare tables of sines. His book ‘Aryabhatta’ deals with Geometry, Mensuration, Progressions, Square root, Cube root and Celestial sphere (spherical Trigonometry). This work, has won him recognition all over the world because of its logical and unambiguous presentation of astronomical observations. Aryabhatta was the pioneer to find the correct value of the constant
Circumference up to four decimals as with respect to a circle Diameter 3.1416. he found the approximate value of and indirectly suggested that is an irrational number. His observations and conclusions are very useful and relevant today. Greek Mathematician Ptolemy, Father of Trigonometry proved the equation sin2A + cos2A = 1 using geometry involving a relationship between the chords of a circle. But ancient Indian used simple algebra to calculate sin A and cos A and proved this relation. Brahmagupta was the first to use algebra in trigonometry. Bhaskarcharaya II (1114 A.D.) was very brilliant and most popular Mathematician. His work known as Siddhantasiron mani is divided into four parts, one of which is Goladhyaya’s spherical trigonometry.
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MATHEMATICS
BASE, PERPENDICULAR AND HYPOTENUSE OF A RIGHT TRIANGLE In ABC, if B = 900, then : (i) For A, we have : (ii)
Base = AB, Perpendicular = BC and Hypotenuse = AC. For C, we have : Base = BC, Perpendicular = AB and Hypotenuse = AC.
So, in a right angled triangle, for a given angle, (i) The side opposite to the right angle is called hypotenuse. (ii) The side opposite to the right angle is called perpendicular. (iii) The third side (r.e., the side forming the given angle with the hypotenuse), is called base.
TRIGONOMETRICAL RARIOS (T-RATIOS) OF AN ANGLE In ABC, let B = 900 and let A be acute. For A, we have : Base = AB, Perpendicular = BC and Hypotenuse = AC. The T-ratios for A are defined as : (i) (ii) (iii) (iv) (v) (vi)
Perpendicular ( P) BC , written as sin A. Hypotenuse ( H ) AC Base ( B) AB Cosine A = , written as cos A. Hypotenuse ( H ) AC Perpendicular ( P) BC Tangent A = , written as tan A. Base ( B) AB Hypotenuse ( H ) AC , written as cosec A. Cosecant A = Perpendicular ( P) BC Hypotenuse ( H ) AC , written as sec A. Secant A = Base ( B) AB Base ( B) AB , written as cot A. Cotangent A = Perpendicular ( P) BC Sine A =
Thus, there are six trigonometrical ratios based on the three sides of a right angled triangle. Aid to Memory : The sine, cosine, and tangent ratios in a right triangle can be remembered by representing them as strings of letters, as in SOH-CAH-TOA. Sine = Opposite Hypotenuse Cosine = Adjacent Hypotenuse Tangent = Opposite Adjacent The memorization of this mnemonic can be aided by expanding it into a phrase, such as “ Some Officers Have Curly Auburn Hair Till Old Age”.
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RECIPROCAL RELATIONS Clearly, we have : (i) cos ec
1 sin
(ii) sec
1 cos
(iii) cot
1 tan
Thus, we have : (i) sin cosec = 1
(ii) cos sec = 1
(iii) tan cot = 1
QUOTAENT RELATIONS
Consider a right angled triangle in which for an acute angle , we have :
Perpendicular P Base B : cos = Hypotenuse H Hypotenuse H P sin H P H P Now, = tan (by def.) B H B B cos H B cos H B H B and, = cos (by def.) P H P P sin H sin cos Thus, tan = and cot = cos sin sin =
POWER OF T-RATIOS
We denote : (i) (sin )2 by sin2 ; and so on. REMARK :
(i)
(ii)
(iii) Ex.1 Sol.
(ii) (cos )2 by cos2 ; (iii) (sin )3 by sin3 ; (iv) (cos )3 by sin3 ; The symbol sin A is used as an abbreviation for ‘the sine of the angle A’. Sin A is not the product of ‘sin’ and A. ‘sin’ separated from A has no meaning. Similarly, cos A is not the product of ‘cos’ and A. similar interpretations follow for other trigonometric ratios also. We may write sin2 A, cos2 A, etc., in place of (sin A)2, (cos A)2, etc., respectively. But cosec A = (sin A)-1 sin-1 A (it is called sine inverse A). sin-1 A has a different meaning, which will be discussed in higher classes. Similar conventions hold for the other trigonometric ratios as well. Since the hypotenuse is the longest side in a right triangle, the value of sin A or cos A is always less than 1 (or, in particular, equal to 1).
Using the information given in fig. write the values of all trigonometric ratios of angle C. Using the definition of t-ratios,
AB 8 4 ; AC 10 5 BC 6 3 cotC = ; AB 8 4
SinC =
BC 6 3 AC 10 5 AC 10 5 tanC = and BC 6 3
cosC =
AB 6 4 ; BC 8 3 AC 10 5 cosC = AB 8 4 tanC =
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3 . find the remaining trigonometric ratios of A. 4
Ex.2
In a right ABC, if A is acute and tan A =
Sol.
Consider a ABC in which B = 900 For A, we have : Base = AB, Perpendicular = BC and Hypotenuse = AC.
Perpendicular 3 Base 4 BC 3 AB 4
tan A =
Let, BC = 3x units and AB = 4x units. Then,
AC =
AB 2 BC 2
=
4 x 2 3x 2
=
25x 2
= 5x units.
Ex.3
In a ABC, right angled at B, if tan A =
Sol.
(i) sinA cosC + cosA sinC We know that tanA =
(ii) cosA cosC – sinA sinC.
[NCERT]
BC 1 AB 3
BC : AB = 1 :
Let BC = k and AB =
Then,
1 , find the value of 3
AC = =
3 3k AB 2 BC 2 …(Phythagoras theorem)
( 3k ) 2 (k ) 2 3k 2 k 2
4k 2 = 2k BC k 1 sinA = AC 2k 2 =
Now,
cosA =
AB AC BC cosC = AC sinC =
and
AB 3k 3 AC 2k 2 3k 3 2k 2 BC k 1 AC 2k 2
(i)
sinA cosC + cosA sinC =
(ii)
cosA cosC – sinA sinC =
1 1 3 3 1 3 , . 1 2 2 2 2 4 4 31 1 3 3 3 . 0 2 2 2 2 4 4 4
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1 2 tan A , verify that 2 sinA cosA = 2 1 tan 2 A
Ex.4
If sinA =
Sol.
We know that SinA =
BC 1 AC 2
Let
BC = k and AC = 2k
AB = =
Now and Now
AC 2 AB 2
(2k ) 2 k 2 4k 2 k 2 3k 2 3k
AB AC BC cosA = AB cosA =
Sol.
1 3 3 …(i) 2 2 2 1 2 2 3 3 3 2 3 3 2 1 3 3 4 2 1 1 3 4 3
2 sinA cosA = 2. .
2 tan A and 1 tan 2 A Ex.5
3k 3 2k 2 k 1 3k 3
2 1
…(ii)
In PQR, right angled at Q, PR + QR = 25 cm and PQ = 5 cm. Find the value of sinP, cosP and tanP. [NCERT] We are given PR + QR = 25 cm PR = (25 – QR) cm By Pythagoras theorem, PR2 = QR2 + PQ2 or (25 – QR)2 = QR2 + 52 or 625 + QR2 – 50 QR = QR2 + 25 or 50QR = 625 – 25 = 600 QR = 12 cm. and PR = (25 – 12) cm = 13 cm Now
and
QR 12 PR 13 PQ 5 cosP = PR 13 QR 12 tanP = PQ 5 sinP =
5
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Ex.6 Sol.
If A and Q are acute angles such that sin B = sinQ, then prove that B = Q. Consider two right ABC and PQR such that sin B = sinQ. We have, sinB =
[NCERT]
AC PR and, sinQ = PQ AB
sinB = sinQ
AC PR AB PQ
AC AB k, (say) PR PQ
…(i)
AC = k PR and AB = k PQ …(ii) Using Pythagoras theorem in triangles ABC and PQR, we have AB2 = AC2 + BC2 and PQ2 = PR2 + QR2
BC =
AB 2 AC 2 and QR =
PQ 2 PR 2
BC QR
AB 2 AC 2
k 2 PQ 2 k 2 PR 2
2 2 BC k PQ PR =k QR PQ 2 PR 2
PQ 2 PR 2
=
BC QR
PQ 2 PR 2
[Using (ii)]
From (i) and (ii), we have,
TRIGONOMETRICAL RATIO OF STANDARD ANGLES T-Ratios of 450 Consider a ABC in which B = 900 and A = 450 Then, clearly, C = 450. AB = BC = a (say). AC =
AB 2 CB 2 a 2 a 2 2a 2 2a.
1 BC a : AC 2a 2 1 AB a cos 450 = : AC 2a 2 BC a tan 450 = 1 AC a 1 1 1 cosec 450 = 2 ; sec 450 = 2 ; cot 450 = 1 0 0 sin 45 cos 45 tan 450 sin 450 =
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T-Ratios of 600 and 300 Draw an equilateral ABC with each side = 2a. Then, A = B = C = 600. From A, draw AD BC. Then, BD = DC = a, BAD = 300 and ADB = 900. Also, AD =
AB 2 BD 2
4a 2 a 2 =
=
3a 2 3a .
T-Ratios of 600 In ADB we have : ADB = 900 and ABD = 600. Base = BC = a, Perp. = AD =
3a and Hyp. AB = 2a.
3a 3 AD : AB 2a 2 BD a 1 cos 600 = : AB 2a 2 3a AD 3 tan 600 = BD a 1 1 2 1 1 cosec 600 = ; sec 600 = 2 ; cot 600 = 0 0 0 sin 60 tan 60 cos 60 3 3 sin 600 =
T-Ratios of 300 In ADB we have : ADB = 900 and ABD = 300.
Base = AD = 3a , Perp. = BD = a and Hyp. AB = 2a.
BD a 1 AB 2a 2 AD 3a cos 300 = 2a AB BD a tan 300 = AD 3a 1 cosec 300 = 2 sin 30 0 sin 300 =
:
3 : 2 1 3 ; sec 300 =
1 1 2 ; cot 300 = 3 0 tan 30 0 cos 30 3
T-Ratios of 00 and 900 T-Ratios of 00 We shall see what happens to the trigonometric ratios of angle A, if it is made smaller and smaller in the right triangle ABC (see figure), till it becomes zero. As A gets smaller and smaller, the length of the side BC decreases. The point C gets closer to point B, and finally when A becomes very close to 00, AC becomes almost the same as AB.
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BC is very close to O. Also AC AB is very close to 1. when A is very close to 00, AC is nearly same as AB and so the value of cos A = AC
When A is very close to 00, BC gets very close to 0 and so the value of sin A =
This helps us to see how we can define the values of sin A cos A when A = 00. We define : sin 00 = 0 and cos 00 = 1. Using these, we have :
sin 0 0 0, tan 0 = cos 0 0 1 1 cot 00 = 0 tan 0 0 1 sec 00 = 1 cos 0 0 1 1 and cosec 00 = 0 sin 0 0 0
(not defined)
(not defined)
T-Ratios of 900 Now, we shall see what happens to the trigonometric ratios of A when it is made larger and larger in ABC till it’becomes 900. As A gets larger and larger, C gets smaller and smaller. Therefore, as in the case above, the length of the side AB goes on decreasing. The point A gets closer to point B. Finally when A is very close 900, C becomes very close to 00 and the side AC almost coincides with side BC (see figure).
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When C is very close to 00, A is very close 900, side AC is nearly the same as side BC, and so sin A is very close to 1. Also when A is very close to 900, C is very close 00, and the side AB is nearly zero, so cos A very close to 0. So, we define: sin 900 = 1 and cos 900 = 0. Using these, we have :
sin 90 0 1 , tan 90 = cos 90 0 0 cos 90 0 1 0 cot 900 = sin 90 0 0 1 1 cosec 900 = 1 0 sin 90 1 1 1 sec 900 = cos 90 0 0
(not defined)
0
and
(not defined)
table for T-Ratios of Standard Angles Angle 00 300 Ratio Sin
0
cos
1
tan
0
cot
Not defined
sec
1
cosec
REMARK:
Not defined
1 2 3 2 1 3
450
600
1 2 1 2
3 2 1 2
1
3
3
1
1 3
2 3
2
2
2
2
2 3
900
1 0 Not defined 0 Not defined 1
(i)
As increases from 00 to 900, sin increases from 0 to 1.
(ii) (iii)
As increases from 00 to 900, cos decreases from 1 to 0. As increases from 00 to 900, tan increases from 0 to .
(iv) (v) (vi) (vii) (viii)
1 ,0 0 90 0 is one. sec As cos decreases from 10 to0, increases from 0 to 900. sin and cos can not be greater than one numerically. sec and cosec can not be less than one numerically. tan and cot can have any value. The maximum value of
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COMPETITION WINDOW T-RATIOS OF SOME ANGLES LESS THAN 900 Angle
150
180
10 22 2
Ratio
Sin
Ex.7 Sol.
5 1 4
3 1 2 2
cos
3 1 2 2
10 2 5 1 2 2 4 2
tan
2 3
25 10 5 5
10 2 5 4 5 1 4 52 5
2 1
In ABC, right angled at B, BC = 5 cm, BAC = 300, find the length of the sides AB and AC. We are given BAC = 300, i.e., A = 300 and BC = 5 cm Now
BC AC AC = 2 5 or 10 cm sinA =
or
sin 300 =
or To find AB, we have,
Ex.8 Sol.
1 2 2 2
360
5 AC
or
=
5 1 AC 2
…[ sin 300 =
1 ] 2
AB = cos A AC AB = cos 300 or 10 3 3 AB or = …[ cos 300 = ] 2 AC 2 3 AB = 10 or 5 3 cm 2 Hence, AB = 5 3 cm and AC = 10 cm. In ABC, right angled at C, if AC = 4 cm and AB = 8 cm. Find A and B. We are given, AC = 4 cm and AB = 8 cm Now sinB =
AC 4 1 AB 8 2
But we know that sin 300 = Now A = 900 – B
1 2
B = 300 …[ A + B = 900]
= 900 – 300 = 600 Hence, A = 600 and B = 300.
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Ex.9
Sol.
sin 30 0 tan 450 cos ec60 0 sec 30 0 cos 60 0 cot 450 1 2 1 0 0 0 sin 30 tan 45 cos ec60 2 3 = 0 0 0 2 1 sec 30 cos 60 cot 45 1 3 2
Evaluate :
[NCERT]
32 34 2 3 = 4 32 3 2 3 =
=
3 3 4 3 3 4 (3 3 4) 3 3 4 3 3 4 (3 3 4)
27 16 24 3 43 24 3 27 16 11
Ex.10 Find the value of in each of the following : Sol.
(i) 2 sin 2 = 3 (i) we have,
(ii) 2 cos 3 = 1
2 sin 2 =
sin 2 =
3 3 2
cos 3 =
1 2
cos 3 cos 600 3 600 = 200 we have,
(iii)
3 tan 2 – 3 = 0
sin 2 sin 600 2 600 = 300 we have, 2 cos 3 = 1
(ii)
(iii)
3 tan 2 – 3 = 0 3 tan 2 = 3 3 tan 2 = 3 3 tan 2 tan 600 2 600 = 300
COMPETION WINDOW
(i) (ii) (iii)
USING TRIGONOMETRIC TABLES A Trigonometric Table consists of three parts : A column on the extreme left containing degrees from 00 to 890. Ten column headed by 0’, 6’, 12’, 18’, 24’, 30’, 36’, 42’, 48’, and 54’, Five column of mean differences, headed by 1’, 2’, 3’, 4’, and 5’, The mean differences is added in case of sines, tangents and secants. The mean difference is subtracted in case of cosines, cotangents and cosecants. The method of finding T-ratios of given angles using trigonometric tables, will be clear from the following example : Find the value of sin 430 52’.
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We have, 43052’ = 43028’ + 4’ In the table natural sines, look at the numbers in the row against 430 and in the column headed 48’ as shown below. From Table of x0 0’ 6’ 12’ 18’ 24’ 30’ 36’ 42’ 48’ 54’ 1’ 2’ 3’ 4’ 5 Natural Sines : Mean Differences 430 0.6820 0.6921 8 Now, sin 43048’ = 0.6921 Mean difference for 4’ = 0.0008
[To be added] [See the number in the same row under 4’] sin 430 52’ = [0.6921 + 0 0.0008] = 0.6929 TO FIND THE ANGLE WHEN ITS T-RATIOS IS GIVEN
Find , when sin = 0.7114. From the table, find the angle whose sine is just smaller than 0.7114. We have sin = 0.7114 Sin 450 18’ = 0.7108 Diff. = 0.0006 Mean difference of 6 corresponds to 3’. Required angle = (450 18’ + 3’) = 45021’ Find , when cos = 0.5248 From the table, find the angle whose cosing is just smaller than 0.5248 We have cos = 0.5248 cos 580 18’ = 0.5255 Diff. = 0.0007 And 7 corresponds to 3’. Required angle = 580 18’ + 3’ = 580 21’ TRY OUT THE FOLLOWING 1. 2.
Using tables find the value of : (i) sin 830 12’ (ii) cos 700 17’ Using tables find the value of if : (i) sin = 0.42’ (ii) cos = 0.8092 (vi) cot = 0.1385
(iii) tan 240 14’ (iv) cosec (30.8)0
(v) sec 680 10’ (vi) cot 390 15’
(iii) tan = 2.91 (iv) cosec = 2.8893
(v) sec = 1.2304
ANSWERS 1. 2.
(i) 0.993. (i) 240 50’
(ii) 0.3373 (ii) 350 59’
(iii) 0.4536 (iii) 710 2’
(iv) 1.9530 (iv) 200 15’
(v) 2.6892 (v) 350 38’
(vi) 1.2283 (vi) 820 7’
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T-RATIOS OF COMPLEMENTARY ANGLES Complementary Angles Two angles are said to be complementary, if their sum is 900. Thus, 0 and (900 – ) are complementary angles. T-ratios of Complementary Angles Consider ABC in which B = 900 and A = 0. C = (900 – ). Let AB = x. BC = y and AC = r. When we consider the T-ratios of (900 – ), then Base = BC, Perp. = AB and Hyp. AC =
AB x = cos . AC r BC y cos (900 – ) = = sin . AC r AB x tan (900 – ) = = cot . BC y 1 1 cosec (900 – ) = = sec . 0 sin(90 ) sin 1 1 = cosec . sec (900 – ) = 0 cos(90 ) sin 1 1 = tan . cot (900 – ) = 0 tan(90 ) cot sin (900 – ) =
(i) sin (900 – ) = cos (iv) cosec (900 – ) = sec
(ii) cos (900 – ) = sin (v) sec (900 – ) = cosec
(iii) tan (900 – ) = cot (vi) cot (900 – ) = tan
Aid to memory Add co if that is not there Remove co if that is there Thus we have, sine of (900 – ) = cosine of sine (900 – ) = cos cosine of (900 – ) = sine of cos (900 – ) = sin tangent of (900 – ) = cotangent of tan (900 – ) = cot cotangent of (900 – ) = tangent of cot (900 – ) = tan secant of (900 – ) = cosecant of sec (900 – ) = cosec cosecant of (900 – ) = secant of cosec (900 – ) = sec In other words : sin (angle) = cos (complement) ; cos (angle) = sin (complement) tan (angle) = cot (complement) ; cot (angle) = tan (complement) sec (angle) = cosec (complement) ; cosec (angle) = sec (complement) where complement = 900 – angle
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Ex.11 Without using tables, evaluate :
Sol.
(i)
sin 530 cos 37 0
(i)
sin 530 sin(90 0 37 0 ) cos 37 0 1 cos 37 0 cos 37 0 cos 37 0
(ii)
cos 49 0 sin 410
(iii)
tan 66 0 cot 24 0
cos 530 cos(90 0 410 ) sin 410 1 sin 37 0 sin 410 sin 410 tan 66 0 tan(90 0 24 0 ) cot 24 0 1 (iii) cot 24 0 cot 24 0 cot 24 0 (ii)
REMARK :
(i) (ii)
[ sin (900 – ) cos ] [ cos (900 – ) sin ] [ tan (900 – ) cot ]
The above example suggests that out of the two t-ratios, we convent one is term of the t-ratios of the complement. For uniformity, we usually convert the angle greater than 450 in terms of its complement.
Ex.12 Without using tables, show that (cos 350 cos 550 – sin 350 sin 550) = 0. Sol. LHS = (cos 350 cos 550 – sin 350 sin 550) = [(cos 350 cos 550 – sin (900 – 550) sin (900 – 350)] = (cos 350 cos 550 – cos 550 cos 350) = 0 = RHS. [ sin (900 – ) cos and cos (900 – ) sin ] Ex.13 Express (sin 580 + cosec 850) in terms of trigonometric ratios of angles between 00 and 450. Sol. (sin 580 + cosec 850) = sin (900 – 50) cosec (900 – 50) = (cos 50 + sec 50). Ex.14 If tan 2A = cot (A – 180), where 2A is an acute angle, find the value of A. Sol. We are given, tan 2A = cot (A – 180) or cot (900 – 2A) = cot (A – 180) …[ cot (900 – 2A) = tan 2A] 900 – 2A = A – 180 or A + 2A = 900 – 180 or 3A = 1080 A = 360 Ex.15 Evaluate : Sol.
sec 29 0 2 cot 80 cot 17 0 cot 450 cot 730 cot 82 0 . 0 cos ec61
sec 29 0 2 cot 80 cot 17 0 cot 450 cot 730 cot 82 0 . cos ec610 =
sec 29 0 2 cot 8 0 cot 17 0 (1) cot(90 0 17 0 ) cot(90 0 8 0 ) 0 0 cos ec(90 29 )
sec 29 0 2 cot 80 cot 17 0 tan 17 0 tan 80 . sec 29 0 1 1 = 1 2 cot 8 0 cot 17 0. . 0 cot 17 cot 80
=
[ cot (900 – ) = tan ]
… tan
1 cot 14
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A BC cos , where A, B and C are interior angles of ABC. 2 2
Ex.16 For a triangle ABC, show that sin Sol.
We know that A + B + C = 1800 Thus we have, B + C = 1800 – A or
A BC 0 90 2 2
0 A BC sin sin 90 or 2 2
or
A BC sin cos . 2 2
T-IDENTITIES We know that an equation is called an identity when it is true for all value of the variables involved. Similarly, an equation involving trigonometric ratios of an angle is called a trigonometric identity, if it is true for all values of the angle(s) the angle(s) involved. The three Fundamental Trigonometric Identities are – A (i) cos2 A + sin2 A = 1 ; 00 A 900 2 2 0 0 (ii) 1 + tan A + cosec A = 1 ; 0 A < 90 (iii) 1 + cot2 A + cosec2 A = 1 ; 00 < A 900
(i)
Geometrical Proof : Consider a ABC, right angled at B. Then we have : AB2 + BC2 = AC2 …(i) By Pythagoras theorem 2 2 0 cos A + sin A = 1 ; 0 A 900 Dividing each term of (i) by AC2, we get
C
B
AB 2 BC 2 AC 2 AC 2 AC 2 AC 2 2
i.e.,
2
AB BC AC AC AC AC
2
(cos A)2 + (sin A)2 = 1 …(ii) cos2 A + sin2 A = 1 0 This is true for all A such that 0 A 900 So, this is a trigonometric identity. 1 + tan2 A = sec2 A ; 00 A < 900 Let us now divide (i) by AB2. We get i.e., i.e.,
(ii)
AB 2 BC 2 AC 2 AB 2 AB 2 AB 2 2
or,
2
AC AB BC AB AB AB
2
i.e., 1 + tan2 A = sec2 A …(iii) 0 This equation is true for A = 0 . Since tan A and sec A are not defined for A = 900, so (iii) is true for all A such that 00 A < 900
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(iii)
1 + cot2 A = cosec2 A : 00 < A < 900 Again, let us divide (i) by BC2, we get
AB 2 BC 2 AC 2 BC 2 BC 2 BC 2
AB BC AC BC BC BC
2
2
2
1 + cot2 A = cosec2 A …(iii) Since cosec A and cot A are not defined for A = 00, therefore (iv) is true for all A such that : 00 < A < 900 Using the above trigonometric identities, we can express each trigonometric ratio in terms of the other rigonometric ratios, i.e., if any one of the ratios is known, we can also determine the value of other trigonometric ratios. Fundamental Identities (Results)
sin 2 cos 2 1 sin 2 1 cos 2 cos 2 1 sin 2
1 tan 2 sec 2 sec 2 tan 2 1 tan 2 sec 2 1
1 cot 2 cos ec 2 cos ec 2 cot 2 1 cot 2 cos ec 2 1
To prove Trigonometrical Identities The following methods are to be followed : Method-I : Table the more complicated side of the identity (L.H.S. or R.H.S. as the case may be) and by using suitable trigonometric and algebraic formula prove it equal to the other side. Method-II : When neither side of the identity is in a simple form, simplify the L.H.S. and R.H.S. separately by using suitable formulae (by expressing all the T-ratios occurring in the identity in terms of the sine and cosine and show that the results are equal). Method-III : If the identity to the proved is true, transposing so as to get similar terms on the same side, or cross-multiplication, and using suitable formulae, we get an identity which is true. Ex.17 Express sin A, sec A and tan A in terms of cot A. Sol. We know that
[NCERT]
1 1 1 cos ec A cos ec 2 A 1 cot 2 A 1 1 sin A …(Dividing num. and deno, by sin A) sec A = cos A cos A sin A
sin A
=
1 cos ec A 1 cot 2 A and tan A cot A cot A cot A
Ex.18 Prove Sol.
=
LHS
sec 2 cos ec 2 tan cot =
sec 2 cos ec 2 (1 tan 2 ) (1 cot 2 ) tan 2 cot 2 2 tan cot
= tan cot tan cot = RHS Hence, proved.
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1 tan A cot A 1 sin A = (cos ecA sin A)(sec A cos A) sin A
Ex.19 Prove (cos ecA sin A)(sec A cos A) Sol.
LHS
1 sin 2 A 1 cos 2 A cos 2 A sin 2 A cos A sin A cos A sin A
=
sin A cos A sin 2 A cos 2 A sin A cos A sin A cos A = 2 sin A cos 2 A sin A cos A sin A cos A 1 = RHS = tan A cot A
1 cos A cos A [ sin 2 A cos 2 A 1 ] [ sin 2 A cos 2 A 1 ]
= sin A cos A
[Dividing the numerator and denominator by sin A cos A.]
Hence, proved.
APPLICATIONS OF TRIGONOMETRY Many times, we have to find the height and distances of many objects in real life. We use trigonometry to solve problems, such as finding the height of a tower, height of a flagmast, distance between two objects, where measuring directly is trouble, some and some times impossible. In those cases, we adopt indirect methods which involves solution of right triangles. Thus Trigonometry is very useful in geography, astronomy and navigation. It helps us to prepare maps, determine the position of a landmass in relation to the longitudes and latitudes. Surveyors have made use of this knowledge since ages. Angle of Elevation The angle between the horizontal line drawn through the observer eye and line joining the eye to any object is called the angle of elevation of the object, if the object is at a higher level than the eye i.e., If a horizontal line OX is drawn through O, the eye of the observer, and P is an object in the vertical plane through OX, then if P is above OX, as in fig. XOP is called the angle of elevation or the altitude of P as seen from O.
Angle of Depression The angle between the horizontal line drawn through the observer eye and line joining the eye to any object is called the angle of depression of the object, if the object is at a higher level than the eye i.e., If a horizontal line OX is drawn through O, the eye of the observer, and P is an object in the vertical plane through OX, then if P is above OX, as in fig. XOP is called the angle of depression of P as seen from O.
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REMARK :
1. 2. 3. 4. 5.
The angle of elevation as well as angle of depression are measured with reference to horizontal line. All objects such as towers, mountains etc. shall be considered as linear for mathematical convenience, throughout this section. The height of the observer, is neglected, if it is not given in the problem. Angle of depression of P as seen from O is equal to the angle of elevation of O, as seen from P. i.e., AOP = OPX. To find one side a right angled triangle when another side and an acute angle are given, the hypotenuse also being regarded as a side.
Re quired = a certain T-ratio of the given angle. Given side 6. 7.
The angle of elevation increases as the object moves towards the right of the line of sight. The angle of depression increases as the object moves towards the right of the line of sight.
COMPWRIRION WINDOW BEARING OF A POINT
The true bearing to a point is the angle measured in degrees in a clockwise direction from the north line. We will refer to the true bearing simply as the bearing.
e.g. (i) the bearing of point P is 650 (ii) the bearing of point Q is 3000 A bearing is used to represent the direction of one point relative to another point. e.g., the bearing of A from B is 600. The bearing of B from A is 2400.
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TRY OUT THE FOLLOWING State the bearing of the point P in each of the following diagrams :
ANSWERS (a) 480
(b) 2400
(c) 1400
(d) 2900
Ex.20 An observer 1.5 m tall, is 28 5 m away from a tower 30 m high. Determine the angle of elevation of the top of the tower from his eye. (NCERT) Sol. Let AB be the height of the tower, CD the height of the observer with his eye at the point D, AB = 30 m, CD = 1.5 m. Through D, draw DE║CA than BDE = where is the angle of elevation of the top of the tower from his eye. AC = horizontal distance between the tower and the observer = 28.5 m BE = AB – AE = (30 – 1.5) m = 28.5 m BDE is right triangle at E,
28.5 BE tan tan tan 1 28.5 DE tan 1 tan 450 450 .
then
Required angle of elevation of the tower = = 450. Ex.21 A vertical post casts a shadow 21 m long when the altitude of the sun is 300. Find : (a) the height of the post. (b) the length of the shadow when the altitude of the sun is 600. Sol.
(c) the altitude of the sun when the length of the shadow is 7 3 m. Let AB be the vertical post and its shadow is 21 m when the altitude of the sun is 300. (a) BC = 21 m, ACB = 300, AB = h metres ABC is rt. ,
AB h 1 tan 30 0 BC 21 3
21 7 3 3 7 3 AB = h, Height of the pole = 7 3 3 3
h
(b)
In this case, we have, ACB = 600, BC = x m AB = 7 3 m ABC is rt. , then :
AB tan 60 0 3 BC h 7 3 3 3 x x x = BC, Length of the shadow = 7 m.
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(c)
In this case : AB = h = 7 3 BC = The length of the shadow = 7 3 m when the altitude of the sun is
AB 7 3 tan tan BC 7 3 tan = 1 = tan 450 450
ABC is rt. , then
Altitude of the sun = = 450 Ex.22 A 1.6 m tall girl stands at a distance of 3.2 m from the lamp-post and casts a shadow of 4.8 m on the ground. Find the height of the lamp-post by using (i) trigonometric ratios (ii) property of similar trianges. (NCERT) Sol. Let PQ be the position of the lamp-post whose height is h metres. i.e., PQ = h metres. AB be the position of the tall girl such that AB = 16. m. Let BC be the shadow of AB such that BC = 4.8 m, ACB = PAE = (corr. s)
(i)
In right ABC,
AB 1.6 1 tan tan tan BC 4.8 3 AB = EQ = 1.6 m. Also AE = BQ = 3.2 m PE = PQ – EQ = (h – 1.6) m
PE h 1.6 1 (h 1.6) 1 tan tan AE 3.2 3 3.2 3 3h – 4.8 = 3.2 3h = 4.8 + 3.2 = 8 3h = 8 2 8 The height of the lamp-post = m = 2 m 3 3 In two s ACB and PCQ, we have : CQ = CB + BQ ACB = ACB = (common) = (4.8 + 3.2) m = 8 metres ABC = PQC = 900 ACB ~ PAE (AA simila) AC CB AB BC AB PC CQ PQ CQ PQ BC AB 4.8 1.6 3 1 3h 8 CQ PQ h 8 8 h 3 2 Thus 3h 8 h m h 2 m 8 3 2 Required height of the lamp-post = PQ = h = 2 m 3 In right ABC,
(ii)
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Ex.23 A captain of an airplane flying at an altitude of 1000 metres sights two ships as shown in the figure. If the angle of depressions are 600 and 300, find the distance between the ships.
Sol.
Let A be the position of the captain of an airplane flying at the altitude of 1000 metres from the ground. AB = the altitude of the airplane from the ground = 1000 m P and Q be the position of two ships. Let PB = x metres, and BQ = y metres. Required : PQ = Distance between the ships = (x + y) metres. ABP is rt. at B ABQ is rt. at B
AB tan 60 0 PB 1000 1000 3x x 3 1000(1.732) x 577.3 m 3
AB tan 60 0 BQ 1000 1 y 1000 3 y 3
Required distance between the ships = (x + y) metres = (577.3 + 1732) m = 2309.3 m Ex.24 Two poles of equal heights are standing opposite to each other on either side of a road, which is 80 metres wide. From a point between them on the road, the angles of elevation of their top are 300 and 600. Find the position of the point and also the height of the poles. Sol. Let AB and CD be two poles of equal height standing opposite to each of their on either side of the road BD. AB = CD = h metres. Let P be the observation point on the road BD. The angles of elevation of their top are 300 and 600. APB = 300, CPD = 600 The width of the road = BD = 80 m, let PD = x metres Then BP = (80 - x) metres Consider right CDP, we have :
CD h tan 60 0 3 h 3 x …(i) PD x In right ABP, we have : AB h 1 80 x h …(ii) tan 30 0 BP 80 x 3 3 h 3x (80 x) 3x 4 x 80 x 20 From (i) and (ii), we get : 80 x h 3 Height of each pole = AB = CD = 3 . x = 20. 3 = 20 (1.732) = 34.64 metres. Position of point P is 20 m from the first and 60 m from the second pole. i.e., position of the point P is 20 m from either of the poles.
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SYNOPSIS
1.
In a right triangle ABC, with right angle B,
sin A
Perpendicular Hypotenuse
cos A
Hypotenuse Perpendicular
cos A
Base Hypotenuse
sec A
Hypotenuse Base
tan A
Perpendicular Base
cot A
Base Perpendicular
1 1 1 : sec A : tan A sin A cos A cot A
2.
cos
3.
The value of sinA or cosA never exceeds 1, whereas the value of secA or cosecA is always greater than or equal to 1.
4.
sin (90 0 A) cos A
cot (90 0 A) tan A
cos (90 0 A) sin A
sec (90 0 A) cos ecA
tan (90 0 A) cot A
cos ec (90 0 A) sec A
5.
sin 2 A cos 2 A 1 : 1 tan 2 A sec 2 : 1 cot 2 A cos ec 2 A
6.
If one of the sides and any other part (either an acute angle or any side) of a right triangle is known, the remaining sides and angles of the triangle can be easily determined.
7.
In a right triangle, the side opposite to 300 is half the side of the hypotenuse.
8.
In a right triangle, the side opposite to 600 is
9.
(i)
3 times the side of the hypotenuse. 2
The line of sight is the line drawn from the eye of an observer to the point in the object viewed by the observer.
(ii)
The angle of elevation of an object viewed is the angle formed by the line of sight with the horizontal when it is above the horizontal level i.e., the case when we raise our head to look at the object.
(iii)
The angle of depression of an object viewed is the angle formed by the line of sight with the horizontal when it is above the horizontal level i.e., the case when we raise our head to look at the object.
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EXERCISE – 1
(FOR SCHOOL/BOARD EXAMS) OBJECTIVE TYPE QUESTIONS
1.
CHOOSE THE CORRECT ONE In ABC , B 90 0 . If AB = 14 cm and AC = 50 cm then tan A equals : (A)
2.
3.
4.
5.
24 25
24 7
(C)
7 24
12 2 cos 3 tan then the value of the is : 13 sin tan sin 5 12 259 (A) (B) (C) 5 102 3 2 2 p q p sin q cos If sec then the value of the is : q p sin q cos p p2 p2 q2 (B) 2 (C) 2 (A) q q p q2 8 If angle A is acute and cos A = then cot A is : 17 8 17 15 (B) (C) (A) 15 8 8 sec is equal to – 1
1 cot 2 cot
(B)
1 cos 2 0
0
sin 30 + cos 60 equals : (A)
1 3 2
(C)
3
(B) 2
0
2
cot 1 cot 2
(C) 1 0
2
The value of 2 tan 60 – 4 cos 45 – 3 sec 30 is : (A) 0 (B) 1 (C) 12
8.
3 The value of tan2 300 – 3 sin2 600 + 3 cosec2 450 is 4
(A) 1 (B) 8 7 sin2 + 3 cos2 = 4 then : (A) tan
10. 11. 12.
25 24
(D)
259 65
(D)
p2 q2 p2 q2
(D)
17 15
1 2
(B) tan
(D) None of these (D) 8
(C) 0
1 2
The solution of the trigonometric equation
(C) tan
(D) 12
1 3
(B)
3 2
(D) tan
cos 2 3,0 0 90 0 : 2 2 cot cos 0 (C) = 60 (D) = 900
(B) = 300 (A) = 00 If cot + cos = p and cot = q, then the value of p2 – q2 is : (B) 4 pq (C) 2 pq (A) 2 pq The value of sin2 150 + sin2 300 + sin2 450 + sin2 600 + sin2 750 is : (A) 1
(D)
cos ec 2 1 cos ec
0
7.
9.
(D)
If sin
(A) 6.
(B)
(C)
5 2
1 3
(D) 4 pq (D) 3
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13.
sin 29 0 sin 610 The value of is : cos 610 cos 29 0 (A) Zero
14. 15.
(B) 1
17. 18. 19.
(B) Zero
21.
22.
23.
24.
25.
29 61 0
(C)
1 2
(D) 2
A flagstaff 6 metres high throws shadow 2 3 metres long on the ground. The angle of elevation is : (B) 450 (C) 900 (D) 600 (A) 300 An observer 3 m tall is 3 m away from the pole 2 3 m high. The angle of the top of elevation of the top from the pole is : (A) 450 (B) 300 (C) 600 (D) 150 An observer 1.5 m tall is 28.5 m away from. a chimney. The angle of elevation of the top of the chimney from her eyes is 450. The height of the chimney is : (A) 30 m (B) 27 m (C) 28.5 m (D) None of these The angle of elevation of the top of a tower from a distance 100 m from its foot is 300. The height of the tower is : (A) 1000 3 m
20.
(D)
The values of x and y which make the following solutions true are: cosx0 = sin 52 and cos y0 = sin (y0 + 10) (A) x = 520, y = 300 (B) x = 380, y = 400 (C) x = 480, y = 520 (D) x = 400, y = 500 If 90 0 and 2 then cos 2 sin 2 equal : (A) 1
16.
61 29
(C)
(B)
200 m 3
(C) 5 3 m
(D)
100 m 3
A kite is flying at a height of 60 m above the ground. The sting attached to the kite is temporarily the to a point on the ground. The inclination of the string with the ground is 600. The length of the staring is : (B) 30 m (C) 20 3 m (D) 60 3 m (A) 40 3 m A tree is broken by the wind. Its top struck the ground at an angle 300 at a distance of 30 m from its foot. The whole height of the tree is : (B) 20 3 m (C) 40 3 m (D) 30 3 m (A) 10 3 m From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are 300 and 450 respectively. If the bridge is at a height of 3 m from the banks then the width of the river is : (A) 3 ( 3 1) m (B) 3 ( 3 1) m (C) ( 3 3) m (D) ( 3 3) m The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. The height of the tower is : (B) 13 m (C) 6 m (D) 2.25 m (A) 5 m A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angles of elevation from his eyes to the top of the building increases from 30 to 600 as he walks towards the building. The distance he walked towards the building is : (B) 57 3 m (C) 38 3 m (D) 18 3 m (A) 19 3 m As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 300 and 600. if one strip is exactly behind the other on the same side of the light-house then the distance between the two ships is : (B) 75 3 m (C) 50 3 m (D) None of these (A) 25 3 m (OBJECTIVE)
Que. Ans. Que. Ans. Que. Ans.
1 B 11 B 2 D
EXERCISE
ANSWER KEY 2 C 12 C 122 B
3 C 13 A 23 C
4 A 14 B 24 A
5 B 15 C 25 C
6 C 16 D
7 A 17 B
8 C 18 A
9 D 19 D
10 C 20 A
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EXERCISE – 2
(FOR SCHOOL/BOARD EXAMS) SUBJECTIVE TYPE QUESTIONS
VERY SHORT ANSWER TYPE QUESTIONS 1.
In the adjoining fig, determine : sin sin
2.
If 5 tan = 4, find the value of
3.
If A = 300. verify sin 2A = 2 sin A cos A :
4. 5. 6. 7.
8. 9. 10. 11. 12.
5 sin 3 cos : sin cos
1 cos ec 2 sec 2 , what is the value of ? cos ec 2 sec 2 5 1 What is the maximum value of ? sec tan What is the value of if sin cos 2 In the given fig ABC is right at B such that AB = 3 cm and AC = 6 cm. Determine ACB. Given that tan
2 tan 30 0 1 tan 2 30 0 cos sin Evaluate : 0 sin(90 ) cos(90 0 ) Evaluate : sin 250 cos 650 cos 250 sin 650 3 If tan A = and A + B = 900, then what is the value of cot B? 4 0 Evaluate :
If tan A = cot B, prove that A + B = 90
2 tan 80 0 3 cot 10 0
13.
Evaluate :
14. 15.
The height of a tower is 10 m. Calculate the height of its shadow when sun’s altitude is 450. What is the angle of elevation of the sum when the length of the shadow of a pole is 3 times of the height of the pole ?
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SHORT ANSWER TYPE QUESTIONS 1. 2.
3. 4. 5. 6.
3 5 sin A 3 sec A 3 tan A . evaluate 5 4 cot A 4 cos ecA 5 cos A q cos ec cot Given cos , find cos ec cot p2 q2
If cos A =
4 12 and cos find sin cos cos sin 5 13 1 1 , prove that sec tan 2 x or If sec x . 4x 2x 1 cot 1 cos 16 If , find . 1 cot 1 cos 9 cos ec sec sec tan 36 If , find . sec tan 49 cos ec sec If sin
7.
If 2 sin + cos = 2, find sin
8.
If
9.
(i)
10. 11.
(ii) If tan (2A + B) = 3 and cot (3A – B) = 3 , find A and B. Find x if : (i) cos (5x – 400) = sin 300. (ii) cosec (x + 300) = cot 450. If A = 600. B = 300, verify each of the following :
12.
1 sin x 1 cos x 7 4 3 . find the value of . 1 sin x 1 cos x If sin (A + B) = 1 and cos (A – B) = 1, find A and B.
(ii)
cot A cot B 1 cot A cot B
cos (A - B) cos A cos B + sin A sin B.
(i)
Assume that tan (A + B) =
(ii)
Assume cos (A - B) = cos A cos B + sin A sin B, find cos 150 when A = 450, B = 300.
tan A tan B , find tan 750 when A = 450, B = 300. 1 tan A tan B
13.
Assume that sin (A + B) = cos A cos B + cos A sin B, if sin A =
14.
Find the value of in each of the following if : (i) 2 cos 3 = 1
15. 16.
cos (A + B)
(i)
(ii) 2 3 tan = 6
(iii)
1 1 , sin B = . Then find angle (A + B) 5 10
1 tan 2 1 1 tan 2 2
Prove that : (i) sin 600 cos 300 + cos 600 sin 300 = sin 900 (ii) cos 900 = 4 cos3cos 300 – 3 cos 300 Evaluate the following :
3 2 0 3 cot 30 3 sin 2 60 0 2 cos ec 2 60 0 tan 2 30 0 4 4 2 0 1 cot 2 60 0 1 tan 30 2 0 2 0 2 0 (ii) ec cos 60 cos 45 sin 45 1 cot 2 60 0 1 tan 2 30 0 1 (iii) cos 2 30 0 cos 2 450 4 sec 2 60 0 cos 2 90 0 2 tan 2 60 0 2 4 0 4 0 2 (iv) 4(sin 30 cos 60 ) 3(cos 45 0 sin 2 90 0 )
(i)
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17.
KUMAR Find the value of x in each of the following :
MATHEMATICS
27 cos 2 450 . (i) 2 x tan 60 3 x sin 30 4 sin 2 60 0 (ii) ( x 1)(sin 4 60 0 cos 4 30 0 ) x(tan 2 60 0 tan 2 450 ) ( x 2) cos 2 450 1 . (iii) ( x 4) sin 2 60 0 ( x 5) tan 2 30 0 x sin 45 0 cos 450 0 . (iv) tan x sin 45 0 cos 45 0 sin 30 0 . (v) sin 2 x sin 60 0 cos 30 0 cos 60 0 sin 30 0 . (vi) tan 3 x sin 450 cos 450 sin 30 0 . 2
0
2
0
18.
If (sec x – 1) (sec x + 1) = 3 then find the value of x.
19.
tan 3 cot 3 sec cos ec 2 sin cos Prove : 1 tan 2 1 cot 2
20.
Prove : (tan cot sec) (tan cot sec ) cos ec 2 .
21.
Prove : (sec 2 A tan 2 A) (cos ec 2 A cot 2 A) 1 2 sec 2 A cos ec 2 A . 2
22.
Prove
23.
Prove
24. 25. 26. 27.
Prove Prove Prove Prove
28.
Prove
29.
Prove
30.
Prove
31.
Prove
32.
Prove
33.
Prove
34.
Prove
35.
Prove
36.
Prove
37.
Prove
38.
Prove
1 cot 2 1 cot : 1 tan 2 1 tan tan cot sin cos . : 2 2 (1 tan ) (1 cot 2 ) : (1 cos sin ) (1 cos sin ) 2 sin (1 sin ) : sin A (1 tan A) cos A (1 cot A) sec A cos ecA . : sec (1 sin ) (sec tan ) 1 : sec (1 sin ) (sec tan ) 1 : cos 2 (1 tan 2 ) sin 2 (1 cot 2 ) 2 cot : 1 3 cos cos ec sin cos : (sec cos ec ) (sin cos ) sec cos ec 2 2(1 sin 2 ) : (tan sec ) 2 (tan sec ) 2 cos 2 (sin cos ) 2 (sin cos ) 2 2 sin cos : (sin cos ) 2 (sin cos ) 2 : (1 sin cos ) 2 2 (1 sin ) (1 cos ) : sin 8 cec 8 (sin 2 cos 2 ) (1 2 sin 2 cos 2 ) : sec 6 tan 6 1 3 sec 2 tan 2 : sec 4 tan 4 2 sec 2 1 cos 3 sin 3 cos 3 sin 3 : 2 sin cos cos sin cos sin sin A cos A sin A cos A 2 : 2 sin A cos A sin A cos A sin A cos 2 A
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39.
Prove :
40.
Prove :
41.
Prove :
42.
Prove :
43.
Prove :
44.
Prove :
45.
Prove :
46.
Prove :
47.
Prove :
48.
Prove :
49.
Prove :
50.
Prove :
51.
Prove :
52.
Prove :
53.
Prove :
54.
Prove :
55.
Prove :
56.
Prove :
57.
Prove :
1 1 2 sec 2 1 sin 1 sin 1 1 2 sec sec tan sec tan cos cos 2 sec 1 sin 1 sin 1 1 1 1 cos ec cot sin sin cos ec cot 1 cos sin 2 cos ec sin 1 cos sin cos sin cos 2 2 sec 2 sin cos sin cos sin 2 cos 2 tan 2 1 sin A sin B cos A cos B 0 cos A cos B sin sin B 1 cot 1 cot 2 2 1 cot 1 cot sin cos 2 cos ec 1 cos ec 1 2 sec cos ec 1 cos ec 1 sec 1 tan tan sec 1 2 sec sec 1 tan tan sec 1 1 cos cos ec cot 1 cos 1 sin cot 2 1 sin (cos ec 1) 2 sec 1 tan sec 1 sec 1 sec tan sec tan sec tan 1 sin 1 2 tan 2 2 tan sec 1 sin cos ec cot 1 2 cot 2 2 cos ec cot cos ec cot cos ec cot 1 cos ec cot cos ec cot sec 1 sec 1 2 cos ec sec 1 sec 1 cos ec cot cos ec cot 2 cos ec cos ec cot cos ec cot
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58. 59. 60. 61. 62. 63. 64. 65. 66. 67.
68.
tan A sec A 1 1 sin A 1 sin A sec A tan A tan A sec A 1 cos A 1 sin A 1 cos A sin A 1 sin A 1 sin A Prove : sec A tan A 1 cos A sin A cos A 1 sin A cos A 1 sin A 1 cos A 1 cos A Prove : cos ecA cot A cos A 1 sin A sin A 1 cos A What is the angle of elevation of a vertical flagstaff of height 100 3 m from a point 100 m from its foot. Prove :
A ladder makes an angle of 600 with the floor and its lower end is 20 m from the wall. Find the length of the ladder. The shadow of a building is 100 m long when the angle of elevation of the sun is 600. Find the height of the building. A ladder 20 m long is placed against a vertical wall of height 10 metres. Find the distance between the foot of the ladder and the wall and also the inclination of the ladder to the horizontal. What is the angle of elevation of the sun when the length of the shadow of the pole is
1 times the height of the 3
pole ? A flagstaff 6 metres high throws a shadow 2 3 metres along on the ground. Find the angle of elevation of the sun. A tree 10 (2 3 ) metres high is broken by the wind at a height 10 3 metres from its root in shch a way that top struck the ground at certain angle and horizontal distance from the root of the tree to the point where the top meets the ground is 10 m. Find the angle of elevation made by the top of the tree with the ground. A tree is broken at certain height and its upper part 9 2 m long not completely separated meet the ground at an angle of 450. Find the height of the tree before it was broken and also find the distance from the root of the tree to the point where the top of the tree meets the ground.
LONG ANSWER TYPE QUESTIONS 1. 2. 3.
The ladder resting against a vertical wall is inclined at an angle of 300 to the ground. The foot of the ladder is 7.5 m from the wall. Find the length of the ladder. A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of vertical pole to the ground. Find the height of the pole if the angle made by the rope with the ground level is 300. The length of a string between a kite and a point on the roof of a building 10 m high is 180 m. If the string makes an angle with the level ground such that tan
4. 5. 6. 7. 8. 9.
4 , how high is the kite from the ground? 3 0
The angle of depression of a ship as seen from the top of 120 m high light house is 60 . How far is the ship from the light house? A boy 1.7 m tall, is 25 m away from a tower and observes the angle of elevation of the top of the tower to be 600. Find the height of the tower. A man 1.8 m tall stands at a distance of 3.6 m from a lamp post and casts a shadow of 5.4 m on the ground. Find the height of the tower. A straight highway leads to the foot of a tower of height 50 m. from the top of the tower, the angles of depression of two cars standing on the highway are 300 and 600. What is the distance between the two cars and how far is each car from the tower ? Two points A and B are on opposite sides of a tower. The top of the tower makes an angle of 300 and 450 at A and B respectively. If the height of the tower is 40 metres, find the distance AB. Two men on either side of a tower 60 metres high observe the angle of elevation of the top of the tower to be 450 and 600 respectively. Find the distance between the two men.
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10.
MATHEMATICS
KUMAR Two boats approach a light house in the middle of the sea from opposite directions. The angles of elevation of the top of the light house from two boats are and . If the distance between the two boats is x metres, prove that the height of the light house is
x cot cot (i) Find h if 60 0 , 45 0 and x = 250 m h
11.
12.
13.
14.
15. 16. 17.
(ii) Find h if 60 0 , 30 0 and x = 400 m A boy standing on a horizontal plane finds a bird flying a distance of 100 m from him at an elevation of 300. A girl standing on the roof of 20 metres high building, finds the angle of elevation of the same bird to be 450. Both the boy and the girl are on opposite sides of the bird. Find the distance of the bird from the girl. Two pillars of equal height stand on either side of a roadway which is 180 metres wide. The angle of elevation of the top of the pillars are 600 and 300 at a point on the roadway between the pillars. Find the height of the pillars and the position of the point. Two lamp posts are 60 metres apart, and the height of the one is double that of the other. From the middle of the line joining their feet, an observer finds the angular elevation of their top to be complementary. Find the height of each lamp. Two lamp posts are of equal height. A boy measured-the elevation of the top of each lamp-post from the midpoint of the line-segment joining the feet of lamp-post as 300. After walking 15 m towards one of them, he measured the elevation of its top at the point where he stands as 600. Determine the height of each lamp-post and the distance between them. When the sun’s altitude increases from 300 to 600, the length of the shadow of a tower decreases by 100 metres. Find the height of the tower. The angle of elevation of the top of a tower from two points at distances a and b metres from the base and in the same straight lien with it are and respectively. Prove that the height of the tower is : From the top of a church spire 96 m high, the angels of depression of two cars on a road, at the same level as the base of the spire and on the same side of it are and where tan
18.
1 1 and tan Calculate the distance 7 4
between tow cars. At a point on the level ground, the angle of the elevation of a vertical tower is found to be such that its tangent is
5 3 . On waling 192 m towards the towards, the tangent of the angel is found to be . Find the height of the 12 4 19.
20.
21.
22.
tower. AB is a straight road leading to C, the foot of a tower, A being at a distance of 120 m from C and B being 75 m nearer. It the angle of elevation of the tower at B be the double of the angle of elevation of the tower at A, find the height of the tower. An aeroplane is observed at the same time by two anti-aircraft batteries distant 6000 m apart to be at elevation of 300 and 450 respectively. Assuming that the aeroplane is traveling directly towards the two batteries, find its height and its horizontal distance from the nearer battery. AT.V tower stands vertically on a bank of canal. From a point on the other bank directly opposite the tower the angle of elevation of the top of the tower is 600. From a point 20 m away from this point on the same bank, the angle of elevation of the top of the tower is 300. Find the height of the tower and the width of the canal. A car is traveling on a straight road leading to a tower. From a point at a distance of 500 m from the tower as seen by the driver is 300. After driving towards the tower for 10 seconds, the angle of elevation of the top of the tower as seen by the driver is found to be 600. Find the speed of the car.
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23.
24. 25.
26.
The height of a hill is 3300 metres. From a point P on the ground the angle of elevation of the top of the hill is 600. A balloon is moving with constant speed vertically upwards from P. After 5 minutes of its movement, a person sitting in it observes the angle of elevation of the top of the hill as 300. What is the speed of the balloon ? A man in a boat rowing away from a light-house 100 m high, takes 2 minutes to change the angle of elevation of the top of the light-house from 600 to 450. Find speed of the boat. From a point on the ground 40 m away from the foot of tower, the angle of elevation of the top of the tower is 300. The angle of elevation to the top of a water tank (on the top of the tower) is 450. Find (i) The height of the tower (ii) The depth of the tank. At a point on a level plane, a tower subtends an angle and a man h metres high on its top an angle . Prove that the height of the tower is :
27.
28. 29.
30.
31. 32.
33. 34. 35.
36. 37.
38.
39.
h tan tan( ) tan
A vertical tower stands on a horizontal plane and is surmounted by a flagstaff of height 12 metres. At a point on the plane, the angle of elevation of the bottom of the flagstaff is 450 and of the top of the flagstaff is 600. Determine the height of the tower: The angles of elevation of the top and the bottom of a flagstaff fixed on a wall are 450 and 300 to a man standing on the other end of the road 20 metres wide. Find the height of the flagstaff and the height of the wall. An aeroplane when flying at a height of 4000 m from the ground passes vertically above another aeroplane at an instant when the angles of the elevation of the two planes from the same point on the ground are 600 and 450 respectively. Find the vertical distance between the aeroplane at that instant. An aeroplane when 6000 metres high passes vertically above another aeroplane at an instant when the angles of the elevation at the same observing point are 600 and 450 respectively. How many metres higher is the one than the other ? Two aeroplane are observed to be in a vertical line. The angle of the upper plane is and a that of the lower is . If the height of the former be H metres, find the height of the latter plane if = 600, = 450, H = 3500 m. The angle of elevation of a Jet fighter from a point A on the ground is 600. After 10 seconds flight, the angle of the of elevation changes to 300. If the Jet is flying at a speed of 432 km/hour, find the height at which the jet is flying. The angle of elevation of a Jet fighter from a point on the ground is 600. After 15 seconds flight, the angle of the of elevation changes to 300. If the Jet is flying at height of 1500 3 m, find the speed of the Jet. From the top of tower 60 metres high, the angle of depression of the top and bottom of a pole are observed to be 450 and 600 respectively. Find the height of the pole and distance of tower from the pole. From the top of a building 60 metres high, the angle of depression of the top and bottom of a vertical lamp-post are observed to be 300 and 600 respectively. Find : (i) The horizontal distance between the building and the lamp-post and (ii) The difference between the height of the building and the lamp-post. From the top of a cliff 200 metres high, the angles of depression of the top and bottom of a tower are observed to be 300 and 600. Find the height of the tower and calculate the distance between them. A man on the deck of a ship is 12 m above water level. He observes that the angle of elevation, of the top of a cliff is 450 and the angle of depression of its base is 300. Calculate the distance of the cliff from the ship and the height of the cliff. From a window (60 metres high above the ground) of a house in a street the angles of elevation and depression of the top and the foot of another house on opposite side of street are 600 and 450 respectively. Show that the height of the opposite house is 60 ( 3 1) metres. A man on the deck of a ship, 16 m above water level, observes that the angle of elevation and depression respectively of the top and bottom of a cliff are 600 and 300. Calculate the distance of the cliff from the ship and the height of the cliff.
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40. 41.
The angle of elevation of a cloud from a point 100 m above a lake is 300 and the angle of depression of its reflection in the lake is 600. Find the height of the cloud. If the angle of elevation of a cloud from a point h metres above a lake be , and the angle of depression of its
tan tan tan tan If the angle of elevation of a cloud from a point h metres above a lake be , and the angle of depression of its
reflection in the lake be , prove that the height of the cloud is : h 42.
reflection in the lake be , prove that the distance (x) of the cloud from the point of observation is :
2h sec . Find x if = 300, = 450 and h = 250 m. tan tan
TRIGONOMETRY
EXERCISE-2 (X)-CBSE
ANSWER KEY
VERY SHORT ANSWER TYPE QUESTION : 1.
5 5 2 2. 4. 5.1 6. 450 7. 300 8. 14 3 3
3 9. 2 10. 1 11.
4 2 13. 14. 10 m 15. 300 3 3
SHORT ANSWER TYPE QUESTION :
1.
5 2. 11
p2 q2 q p q q 2
2
3.
71 63 31 3 5. 6. 7. 1, 8. 3 9. (i) A = B 450 (ii) A = 180, B = 240 5 65 17 97
6 2 13. 450 14. (i) 200 (ii) 600 (iii) 300 4 10 2 16 83 16. (i) (ii) (iii) (iv) 2 17. (i) , (ii) 3, (iii) 8, (iv) 450, (v) 150, (vi) 150 18. 600 61. 600 62. 40 m 3 8 3 3 63. 173.2 m 64. 17.32 m, = 300 65. 600 66. 600 67. 600 68. 9( 2 1) m , 9 m 10. (i) 200 (ii) 600 12. (i) 2 3 (ii)
LONG ANSWER TYPE QUESTION : 1. 8.66 m 2. 10 m 3. 154 m 4. 69.28 m 5. 45 m 6. 3 m 7. 86.5 m ; 57.67 m, 28.83 m 8. 109.28 m 9. 94.64 m 10. (i) 158.5 m (ii) 173.2 m 11. 42.42 m 12. 135 m from one end, h = 77.94 m 13. 21.21 m, 42.42 m 14. Distance = 45m, height = 12.99 m 15. 86.6 m 17. 288 m 18. 180 m 19. 60 m 20. 3000 ( 3 1) m, 3000 ( 3 1) m, 21. Height = 17.32 m, width = 10 m 22. 120 km/hr 23. 26.4 km/hr 24. 1.269 km/hr 25. (i) 23.1 m (ii) 16.91 m 27. 16.392 m 28. 8.45 m, 11.55 m 29. 1690.66 m, 30. 2536 m 31. 2020.78 m 32. 1039.2 m 33. 720 km/hr 34. h = 25.36 m, x = 34.64 m 35. (i) 34.64 m, (ii) 20 m
1 3
36. Height = 133 m, Distance = 115.46 m 37. Height = 32.784 m, Distance = 20.784 m 39. Height = 48 m, Distance = 27.71 m 40. 200 m 42. 1366 m
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EXERCISE – 3
1. 2. 3. 4. 5. 6. 7. 8.
(FOR SCHOOL/BOARD EXAMS)
PREVIOUS YEARS BOARD QUESTIONS SHORT ANSWER TYPE 1 0 0
Without using tables, find the value of 14 sin 30 + 6 cos 60 = 5 tan 450.
[ICSE-2004]
cos ecA cos ecA Prove that : 2 2 tan 2 A cos ecA 1 cos ecA 1 sec . cos ec (90 0 ) tan cot(90 0 ) sin 2 550 sin 2 350 Evaluate : tan 10 0 tan 20 0 tan 60 0 tan 70 0 tan 80 0 0
[CBSE-Al-2004C] [CBSE-Al-2004C]
Without using mathematical tables, find the value of x if cos x = cos 60 cos 300 + sin 600 sin 300. [ICSE-2005]
2 tan 530 cot 80 0 Without using trigonometric tables, evaluate : cot 37 0 tan 10 0 sin 80 0 Without using trigonometric tables, evaluate : sin 59 0 sec 310 0 cos10 sin 250 cos 250 Without using tables, evaluate : sec 650 cos ec650 sin A (cos ecA cot A) Prove the (1 cos A)
[ICSE-2006] [ICSE-2007] [ICSE-2008] [ICSE-2008]
9.
In the fig AD = 4 cm, BD = 3 cm and CB = 12 cm, find cot
10.
11sin 70 0 4 cos 530 cos ec37 0 Without using the trigonometric tables, evaluate the following : 7 cos 20 0 7 7 tan 150 tan 350 tan 550 tan 750
[CBSE-Delhi-2008]
[CBSE-Delhi-2008] 0
11.
Without using the trigonometric tables, evaluate the following :
tan 800 ]
12. 13.
sin 18 3 [tan 10 0 tan 30 0 tan 40 0 tan 50 0 0 cos 72 [CBSE-Delhi-2008] [CBSE-Al-2008]
If sin = cos , find the value of . Without using the trigonometric tables, evaluate the following : (sin 2 25 0 sin 2 65 0 ) 3 (tan 50 tan 15 0 tan
30 0 tan 750 tan 850 ) .
[CBSE-Al-2008]
14.
Without using trigonometric tables, evaluate the following : (cos 25 cos 65 ) cos ec sec(90 ) cot tan(90 ) . [CBSE-Al-2008]
15.
If 7 sin 2 3 cos 2 4 show that tan
16.
If tan A
2
1 . 3
5 , find the value of (sin A + cos A) sec A. 12
0
2
0
[CBSE-Al-2008] [CBSE-Foreign-2008]
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17.
KUMAR If sec 4A = cosec (A – 200), where 4A is an acute angle, find the value of A. OR
In a ABC, right angled at C, if tan A 18. 19. 20. 21. 22.
MATHEMATICS [CBSE-Foreign-2008]
1 , find the value of sin A cos B + cos A sin B. 3
7 , find the value of tan A + cot A. 25 1 If sin = , find the value of [2 cot2 + 2) 3 sin 3 cos 3 sin cos Simplify : sin cos If sec 2 (1 sin ) (1 sin ) k , then find the value of k. (2 2 sin )(1 sin ) 15 , then evaluate If cot 8 (1 cos )(2 2 cos ) If cos A =
[CBSE-Foreign-2008] [CBSE-Delhi-2009] [CBSE-Delhi-2009] [CBSE-Al-2009]
OR Find the value of tan 600 geometrically. 23. 24.
[CBSE-Al-2009]
15 and A + B = 900. find the value of cosec B. [CBSE-Foreign-2009] If sec A = 7 7 cos 70 0 3 cos 550 cos ec350 Without using trigonometric tables, evaluate : 2 sin 20 0 2 2 tan 50 tan 250 tan 450 tan 850 tan 650 [CBSE-Foreign-2009] SHORT ANSWER TYPE II
1. 2.
sec 2 54 0 cot 2 36 0 Evaluate : 2 sin 2 380 sec 2 52 0 sin 2 450 2 0 2 0 cos ec 57 tan 33 Prove that following : (tan A tan B ) 2 (1 tan A tan B ) 2 sec 2 A sec 2 B .
[Al-2005] [Foreign-2005]
OR
sin 150 cos 750 cos 150 sin 750 cos sin(90 ) sin cos(90 ) sec tan 1 cos . Prove that: tan sec 1 1 sin Evaluate : sec 2 10 0 cot 2 80 0
3.
[Foreign-2005]
OR Without using trigonometric tables, evaluate the following :
cot(90 ).sin (90 ) cot 40 0 (cos 2 20 0 cos 2 70 0 ) . sin tan 450
4.
Without using trigonometric tables, evaluate the following :
5.
sec 2 cot 2 (90 ) (sin 2 40 0 sin 2 50 0 ) . 2 0 2 0 cos ec 67 tan 23 Prove that: (1 tan A) 2 (1 tan A) 2 2 sec 2 A . sin tan Prove that: 1 sec . 1 cos
6.
[Delhi-2005C]
[Al-2005C] [ICSE-2005] [ICSE-20056]
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7.
Prove that:
sin cos sin cos 2 sec 2 . sin cos sin cos tan 2 1
[ICSE-20056]
OR
sec (90 ) cot 2 2 cos 2 60 0 tan 2 28 0 tan 2 62 0 . Without using trigonometric tables : 2(sin 2 25 0 sin 2 650 ) 3(sec 2 430 cot 2 47 0 ) 1 1 1 1 Prove that : cos ec cot sin sin cos ec cot 2
8.
[Al-2006]
OR
cos ec (90 ) tan 2 2 tan 2 30 0 sec 2 52 0 sin 2 380 . 4(cos 2 480 cos 2 42 0 ) cos ec 2 70 0 tan 2 20 0 sin 2 sin 2 (90 ) 3 cot 2 30 0 sin 2 54 0 sec 2 36 0 Without using trigonometric tables : . 3(sec 2 610 cot 2 29 0 ) 2(cos ec 2 650 tan 2 250 ) 2
Without using trigonometric tables :
9. 10.
[Foreign-2006]
Without using trigonometric tables evaluate the following :
cos 2 20 0 cos 2 70 0 2 cos ec 2 580 2 cot 580 tan 32 0 4 tan 130 tan 37 0 tan 450 tan 530 tan 77 0 2 0 2 0 sec 50 cot 40 OR Prove that : 11.
sec 1 sec 1 2 cos ec sec 1 sec 1
[Delhi-2006C]
Without using trigonometric tables evaluate the following :
sec 39 0 2 tan 17 0 tan 38 0 tan 60 0 tan 52 0 tan 730 3(sin 2 310 sin 2 59 0 ) 0 cos ec51 3 3 cos 55 0 4(cos 70 0. cos ec 20 0 ) (ii) 7 sin 350 7(tan 5 0. tan 25 0. tan 45 0. tan 65 0. tan 85) cot 54 0 0 0 0 0 0 (iii) tan 7 . tan 23 . tan 60 . tan 67 . tan 83 sin 20 0.sec 70 0 2 0 tan 36 sin A 1 1 cos A Prove that : sin A 1 1 cos A cot A cos A cos ecA 1 Prove that : cot A cos A cos ecA 1 (i)
12. 13.
14.
15.
OR Prove that : (1 + cot A – cosec A) (1 + tan A + sec A) = 2 Prove that : (sin cos ec ) 2 (cos sec ) 2 7 tan 2 cot 2 OR Prove that sin (1 + tan ) + cos (1 + cot ) = sec + cosec Prove that : (1 cot A tan A)(sin A cos A) sin A tan A cot A cos tA . OR
[Al-2006C] [Delhi-2007] [Al-2007] [ICSE-2007]
[CBSE (Delhi)-2008]
[CBSE - Al-2008] [CBSE-foreign-2008]
cos 58 0 cos 38 0 cos ec52 0 . 3 0 0 0 0 sin 32 tan 15 tan 60 tan 75
Without using trigonometric tables evaluate the following : 2
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16.
MATHEMATICS
KUMAR Find the value of sin 300 geometrically. OR Without using trigonometric tables, evaluate :
cos 580 sin 22 cos 380 cos ec52 0 sin 32 0 cos 680 tan 180 tan 350 tan 60 0 tan 72 0 tan 550 [CBSE-Delhi-2009]
17. 18.
1.
2 2 5 cos ec 2 580 cot 580 tan 32 0 tan 130 tan 37 0 tan 450 tan 530 tan 77 0 . [CBSE-Al-2009] 3 3 3 2 4 sin 2 sin Prove that : sec 2 1 [CBSE-foreign-2009] 2 cos 4 cos 2 LONG ANSWER TYPE Evaluate :
On a horizontal plane there is a vertical tower with a pole on the top of the tower. At a point 9 metres away from the foot of the tower the angle of elevation of the top and bottom of the flag pole are 600 and 300 respectively. Find the height of the tower and flag pole mounted on it. OR From a building 60 metres high the angles of depression of the top and bottom of lamp-post are 300 and 600 respectively. Find the distance between lamp-post and building. Also find the difference of height between building and lamp-post. [Delhi-2008]
2.
From the top of a cliff 92 cm high, the angle of depression of a buoy is 200. calculate to the nearest metre, the distance of the bucy from the foot of the cliff. [ICSE-2005]
3.
The shadow of a vertical tower AB on level ground is increased by 10 m, when the altitude of the sun changes from 450 to 300. Find the height of the tower and give your answer correct to
4.
1 of a metre. 10
[ICSE-2006]
The angle of depression of the top and the bottom of a building 50 metres high as observed from the top of a tower are 300 and 600 respectively. Find the height of the tower and also the horizontal distance between the building and the tower. [Delhi-2006] OR The angle of elevation of the top of a tower as observed from a point on the ground is ' ' and on moving ‘a’ metres towards the tower, the angle of elevation is ' ' Prove that the height of the tower is
5.
6.
7.
a tan tan . tan tan
A man on the top of a vertical tower observes a car moving at a uniform speed coming directly tower it. If it takes 12 minutes for the angle of depression to change from 300 to 450 how soon after this, will the car reach the tower? [Al-2006C] A boy standing on a horizontal plane finds at a distance of 100 m from him at an elevation of 300. A girl standing on the roof of 20 metre high building, finds the angle of elevation of the same bird to be 450. Both the boy and the girl are on opposite sides of the bird. Find distance of bird from the girl. [Delhi-2007] Statue 1.46 m tall, stands on the top of the pedestal . From a point on the ground, the angle of elevation of the top of the statue is 600 and from the same point, the angle of elevation of the top of the pedestal is 450. Find the height of the pedestal. (use
3 1.73 )
[CBSE-Delhi-2008]
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A person standing on the bank of a river observes that the angle of elevation of the top of a tree standing on the opposite bank is 600. When he moves 40 m away from the bank, he finds the angle of elevation to be 300. Find the
8.
height of the tree and the width of the river. (use
3 = 1.732)
[CBSE-Delhi-2008]
The angle of elevation of a jet fighter from a point A on the ground is 600. After a fight of 15 seconds, the angle of elevation changes to 300. If the jet is flying at a speed of 720 km/hour, find the constant height at which the jet is
9.
flying. (use
3 = 1.732)
[CBSE-Al-2008]
10.
The angle of elevation of an aeroplane from a point A on the ground is 600. After a flight of 30 seconds, the angel of elevation changed to 300. If the plane is flying at a constant height of 3600 3 m, find the speed in km/hour of the plane. [CBSE-foreign-2008]
11.
A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 300, which is approaching the foot of the tower with a uniform speed. Six seconds later the angle of depression of the car is found to be 600. Find the time taken by the car to reach the foot of the tower from this point. [CBSE-Delhi-2009]
12.
An aeroplane when flying at a height of 3125 m from the ground passes vertically below another plane at an instant when the angels of elevation of the two planes from the same point on the ground are 300 and 600 respectively. Find the distance between the planes at that instant [CBSE-Al-2009]
13.
A man is standing on the deck of a ship which is 25m above water level. He observes the angle of elevation of the top of a lighthouse as 600 and the angle of depression of the base of the light house as 45+0. Calculate the height of the lighthouse. [CBSE-foreign-2009]
TRIGONOMETRY
ANSWER KEY
EXERCISE-2 (X)-CBSE
SHORT ANSWER TYPE QUESTION-I
2 12 17 625 4. 300 5. 1 6. 2 7. 1 9. 10. 1 11. 2 12. 450 13. 2 14. 2 16. 17. 220 or 1 18. 5 12 168 3 625 15 or 3 23. 24. 5 19. 18 20. 1 21. 1 22. 64 7 1 5 3.
SHORT ANSWER TYPE QUESTION-II
1 5 2 25 5 2. or 2 3. or 1 4. 2 7. or 8. or 9. 10. 1 11. (i) 0 (ii) (iii) 2 3 6 7 12 2 3 1 1 16. or 17. – 1 2 3 1.
3 1 15. or 1
LONG ANSWER TYPE QUESTION-II 1. 15.588 m, 5.196 m or 34.64 m and 20 m 2. 253 m 3. 13.66 m 4. 75 m and 43.3 m 5. 16 minutes 23 seconds 6. 30 2 7. 2 m 8. 34.64 m and 20 m 9. 2598 m 10. 864 km/hr 11. 3 seconds 12. 2083.33 m 13. 68.25 m
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EXERCISE – 4
(FOR OLYMPIADS)
CHOOSE THE CORRECT ONE
ax sin by cos ax by a 2 b 2 and 0 then (ax)2/3 + (by)2/3 is equal to : 2 2 cos sin cos sin
1.
If
2.
(A) (a2 – b2)2/3 (B) (a2 + b2)2/3 (C) (a – b)2/3 (D) None of these The sides of a right angled triangle form a geometric progression, find the cosines of the acute angles. (If a, b, c are in G.P. b2 = ac) :
5 1 5 1 5 1 5 1 and (B) and 2 2 2 2 5 1 5 1 (C) and (D) None of these 2 2 1 cos sin 2 sin , then is equal to : If y 1 cos sin 1 sin 1 (B) 1 – y (C) (D) None of these (A) 1+ y y 0 0 (A)
3.
4.
cot 36 cot 72 is equal to : (A)
5.
1 5
(B) 2
0
(C) 1 2
0
2
0
(D) None of these 2
0
The value of cos 15 – cos 30 + cos 45 – cos 60 + cos 75 is : (A) 2
6.
2
1 50
If
(B) 0
(C)
x sin 2 cos and y cos 2 sin , then : 2
(A) ( x y )
2/3
( xy )
2 2/3
1 4
x2 (B) y
1
(D) 2/3
y2 x
2/3
7.
(C) x2 + y2 = x2y2 (D) None of these If x sec tan and y cos ec cot , then xy 1 is equal to : (B) x – y (C) 2x + y (A) x + y
8.
If 5 sin 3 , then (A)
9.
1 4
11.
12.
(B) 4
The value of the expression 1 (A) cos y
10.
sec tan is equal to : sec tan
(B) 1
(C) 2
1 2
1
(D) y – x
(D) None of these
sin 2 y 1 cos y sin y is equal to : 1 cos y sin y 1 cos y (C) 0
(D) sin y
1 If sec x , x R, x 0 , then the value of sec tan is : 4x 1 1 (A) 2x (B) (C) 2x or (D) None of these 2x 2x p sin q cos p is : If tan , then the value of q p sin q cos p2 q2 p2 q2 (B) 2 (C) 0 (D) None of these (A) 2 p q2 p q2 2 2 2 If m = tan sin and n tan sin , then (m – n ) is equal to : 38
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13. 14. 15. 16. 17.
(A) mn (B) 4 mn (C) 16 mn If x cos b sin and y a sin cos then a2 + b2 is equal to : (A) x2 – y2 (B) x2 + y2 (C) (x+ y) If cos
2
(B) 0 (C) – 2 (D) 1 (A) 2 ABC is a triangle, right angled at A. If the length of hypotenuse is 2 2 times the length of perpendicular from A on the hypotenuse, the other angles of the triangle are : (A) 22.50, 67.50 (B) 300, 600 (C) 450, 450 (D) None of these 3 3 If sin A + cos A = m and sin A + cos A = n, then : (B) m3 – 3m + 2n = (A) m3 + 3m + 2n = 0 3 (D) m3 – 3m + n = 0 (C) n – 3n + 2m = 0 If sin 2 3 cos 2 0 , then cos3 sec3 is equal to : (B) 9
20. 21.
3 2 (a 1) 2 4
If 0 x
or
25.
26.
1 4
3 2 (a 1) 2 4
(C)
3 4(a 2 1) 2 4 0
(D)
3 3(a 2 1) 2 4
2
1 , then the values of A and B lying between 00 and 900 are respectively: 20 0 0 0
(B) 60 and 30
(C) 45 15
(D) None of these
and 81sin 2 x 81cos 2 x 30 , then x is equal to : (B)
or 0
(C)
or
(D) None of these
4 2 4 6 If m 2 m '2 2mn' cos 1, n 2 n '2 2nn' cos 1 , and mn m' n'(mn' m' n) cos 0 , then m2 + n2 is 3
equal to : (A) sin 2 24.
(B) 1
If sin( A B ) cos( A B )
(A) 23.
(D)
The quadratic equation whose roots are sin 180 and cos 36 is : (A) 4 x 2 2 5 x 1 0 (B) 4 x 2 2 5 x 1 0 (C) x 2 2 5 x 1 0 (D) 4 x 2 2 5 x 1 0 If cos sec 2 , then the value of cos 2 sec 2 is : (A) 1 (B) 2 (C) 4 (D) None of these
(A) 300 and 600 22.
(C) 4
If sin cos a , then sin 6 cos 6 is equal to : (A) 1
19.
(D) None of these
2
x y y x y sin 1 0 and sin cos 1 0 then 2 2 is equal to : b a b a b
(A) 18 18.
(D) 4 mn
(B) cos ec 2
(C) cos 2
cos A sin A p and q , then tan A is equal to : sin B cos p q2 1 q2 1 p q2 1 (B) (C) (A) q 1 p2 1 p2 q 1 p2 T T5 If Tn sin n cos n , then 3 is equal to : T1 T T7 T T5 T T6 (A) 5 (B) 3 (C) 9 T7 T4 T3
(D) None of these
If
The number of values of which lie between 0 and (A) 1
(B) 2
2
(C) 3
(D) None of these
(D)
T6 T9 T4
and satisfy the equation sin 4 2 sin 2 1 0 is : (D) None of these
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27.
The greatest angle of a cyclic quadrilateral is 3 times least. The circular measure of the least angle is : (A) 600
28.
(B)
30.
31. 32. 33.
6
If 0 < x <
3 1 m 3 1
(B)
(B) 30 (3 3 ) m
2
2
a2 3
3 1 3 1 60 m
(C)
(D) None of these
(C) 30 (3 3 ) m
(D) 30 ( 3 1) m
(C) – 1
(D) None of these
2
(B) 1
, then the largest angle of a triangle whose sides are 1, sin x, cos x is : (B)
2 a2 bc
3
(C)
2
–x
(D) x
(B)
c2 ab
(C)
b2 –x ac
(D) a + b
A rectangle with an area of 9 square metre is inscribed in a triangle ABC having AB = 8 m, BC = 6 m and ABC 90 0 . The dimensions of the rectangle (in metres) are : (A) 2,
3 9 or 6, 2 2
(B) 1 , 9 or 3, 3
(C) 2, 4.5
(D) 4, 2.25
From the top of a light house, the angles of depression of two stations on opposite sides of it at distance ‘a’ apart are and . The height of the light house is : (A)
40.
(D)
ABC is right angled at C, then tan A + tan B = (A)
39.
a2 6
sin cos 3 sin cos is equal to : 6
(A)
38.
(C)
The angles of elevation of the top of a tower as observed from the bottom and top of a building of height 60 m are 600 and 450 respectively. The distance of the base of the tower from the base of the building is :
(A) 0
37.
(D) None of these
3
(A) cos 2 cos 2 (B) tan 2 tan 2 (C) tan 2 tan 2 (D) sec 2 sec 2 From the top of a light house, 60 m high with its base at the sea level, the angle of depression of a boat is 150. The distance of the boat from the foot of the light house is :
(A) 30 ( 3 1) m
36.
If sin x + sin2x = 1, then the value of cos12x + 3cos8x + cos6x + 2cos4x + cos2x – 2 is equal to : (A) 0 (B) 1 (C) 2 (D) sin2x The angles of elevation of the top of a TV tower from three points A, B and C in a straight line (in the horizontal plane) through the foot of tower are , 2 and 3 respectively. If AB = a, the height of tower is : (A) a tan (B) a sin (C) a sin 2 (D) a sin 3 2 2 2 2 2 2 2 The expression cosec A cot A – sec A tan A – (cot A – tan A) (sec A cosec2A – 1) is equal to : (A) 0 (B) 1 (C) – 1 (D) None of these 2 2 (1 tan tan ) (tan tan ) is equal to :
3 1 3 1 60 m
35.
4
(B) 3a2
(A) 34.
(C)
A circle is inscribed in an equilateral triangle of sides a, the area of any square inscribed in the circle is : (A) 6a2
29.
a cot cot
(B)
a cot cot
(C)
a cot cot cot cot
The value of the expression tan 10 tan 20 tan 30……tan 890 is equal to : (A) 0 (B) Not defined (C) 1
(D)
a tan tan cot cot
(D)
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41. 42. 43.
If sin 1 sin 2 sin 3 3 then cos 1 cos 2 cos 3 is equal to : (A) 3 (B) 2 (C) 1 (D) 0 If sinx + sin2x = 1, then cos8x + 2cos6x + cos4x is equal to : (A) 0 (B) – 1 (C) 2 (D) 1 Which of the following is not possible ?
5 1 t 2 (B) cos , t 0 (C) tan 100 1 t 2 7 cot 2 sin cos (0 90 0 ) if equals :
(A) sin 44. 45.
46.
: n n
. n n
49. 50.
: n n
(B) cos
If tan sec 3 , 0 (A)
48.
5 2
(A) 450 and 900 (B) 450 and 600 (C) 450 only (D) 900 only In a triangle ABC right angled at C, tan A and tan B satisfy the equation : (A) abx2 – (a2 + b2) x – ab = 0 (B) abx2 – c2x + ab = 0 2 2 2 (C) c x – abx + c = 0 (D) ax2 – bx + a = 0 The area of the circle and the area of a regular polygon of n sides and of perimeter equal to that of the circle are in the ratio of : (A) tan
47.
(D) sec
(B)
3
2
: n n
(C) sin
(D) cot
, then is equal to : (C)
6
(D) None of these
4
A tower subtends an angle at a point ‘A’ in the plane of it’s base and the angle of depression of the foot of the tower at a height b just above A is B. Then the height of the tower is : (B) b cot tan (C) b tan tan (D) b cot cot (A) b tan cot If sinx + sin2 x = 1, then cos2x + cos4x is equal to : (A) 1 (B) – 1 (C) 2 (D) 0 The angle of elevation of a tower from a point A due south of it is x and from a point b due to east of A is y. if AB = , the height h of the tower is : (A)
(B)
cot y cot x 2
2
tan y tan x 2
2
OBJECTIVE
(C) cot 2 y cot 2 x (D) tan 2 y tan 2 x
EXERCISE -4
ANSWER KEY
Que. Ans.
1 A
2 C
3 D
4 B
5 D
6 B
7 D
8 B
9 A
10 C
11 A
12 C
13 B
14 A
15 A
Que. Ans. Que. Ans. Que. Ans.
16 B 31 A 46 A
17 A 32 D 47 B
18 B 33 A 48 A
19 D 34 D 49 A
20 C 35 B 50 A
21 C 36 A
22 A 37 A
23 B 38 A
24 A 39 B
25 A 40 C
26 D 41 D
27 B 42 D
28 C 43 B
29 D 44 A
30 C 45 B
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COMPETITION WINDOW LAW OF SINES We use the sine rule for non-right angled triangles to find the lengths and angles. In trigonometry, the law of sines (also known as the sines law, sine formula, or sine rule) is an equation relating the lengths of the sides of an arbitrary triangle to the sines of it’s angle. According to the law.
a b c sin A sin B sin C Where a, b and c are the lengths of the sides of a triangle, and A, B and C are the opposite angles. To use the sine rule, choose an appropriate pair, depending on what you know in the triangle e.g.,
b c a b a c or or sin A sin B sin a sin C sin B sin C
If you are finding an angle, you can invert the formulae. e.g., E.g.
sin A sin B sin a sin C sin B sin C or or b c a b a c
Find the length of PQ in triangle PQR. Use the sine rule
p q r sin P sin Q sin R q r Use : sin Q sin R r 165 0 sin 84 sin 230 165 sin 230 r 64.8 metres sin 84 0 1. 2.
TRY OUT THE FOLLOWING Find the length of BC in triangle ABC, if A 780 , C 180 and AB = 26 cm.
[Ans : 82.3 cm]
Find the length of AC in triangle ABC, if B 86 , A 74 and AB = 35 cm. RELATION TO THE CIRCUM RADIUS (R) 0
0
a b c 2R sin A sin B sin C
[Ans : 102.1 m]
…(i)
AREA OF A TRIANGLE For any triangle ABC, the area is given by
A
c
B 1 1 1 bc sin A ca sin B ab sin C 2 2 2
b
a
C …(ii)
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From (i)
sin C =
c 2R
…(iii)
1 ab sin C 2 abc From (iii) and (iv) R 4 From (ii)
…(iv)
LAW OF COSINES In any triangle ABC,
b2 c2 a2 cos A 2bc 2 a c2 b2 cos B 2ac 2 a b2 c2 cos C 2ab
A c B
b C
a
To use the cosine rule, you need to know either two sides and the included angle or all three sides. e.g., Find the length of AC in ABC use :
a2 c2 b2 2ac (4) 2 (4 2 ) 2 AC 2 cos 450 2 4 4 2
cos B
32 = 48 – AC2 AC = 4 cm
A 4 cm
B
4 2 cm
C
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EXERCISE – 5
(FOR IIT-JEE/AIEEE)
CHOOSE THE CORRECT ONE Based On Sine Rule (Q.No. 1-20) 1.
In ABC. AB = 30 cm and C = 450. The length of the radius of circumcircle of ABC is : (B) 5 2 cm (C) 15 3 cm (D) 5 3 cm (A) 15 2 cm
2.
The radius of the circumcircle of ABC is
3.
(A) 300 (B) 600 (C) 90 (D) 45 In ABC, A : B = 1 : 3 : 8. If AB = 10 cm, the length of AC is : [Use : is (180 0 ) = sin ] (A)
4.
5. 6.
10 6 cm 3
(B)
10 3 cm 3
2 3 cm. If BC = 2 cm, the size of angle A is : 3 0 0
(C)
The measure of angle x in the triangle below is :
10 3 cm 6
(B) 57.010 (C) 590 (D) None of these (A) 540 In a circle of radius 7 cm, the are AB subtends an angle of 1200 at the centre. The length of chord AB is : (B) 3 2 cm (C) 5 3 cm (D) 2 3 cm (A) 7 3 cm In a triangle ABC, a = 6, b = 12 and B = 600. The value of sin A is ; (A)
3 cm 4
(B)
1 cm 3
(C)
1 cm 2
2 , then B is equal to : 3 0
7.
In ABC. a = 2, b = 3 and sin A
8.
(A) 300 (B) 600 (C) 90 In ABC. a = 4, c = 12 and C = 600, then the value of sin A is : (A)
9. 10.
1 2 3
cm
(B)
1 cm 2 3
(C)
10 sin 2a sin 3a
(C)
2 cm 3
(D) None of these
(D) 1200 (D)
3 2
In an isosceles triangle ABC, the base AB = 12 cm and the angle at the top is 300. D is a point on the side BC such that CAD : DAB = 1 : 4. The length of the radius of circumcircle of ABC is : (B) 5 2 cm (C) 6 2 cm (D) 10 2 cm (A) 3 2 cm The base of an isosceles triangle is 10 cm, and the angle at the base is 2a. the length of the angle bisector of one of the base angles is : [Use : sin(180 0 ) sin ] (A) 10 sin 2a cos 2a
11.
(D) None of these
(B)
10 sin 3a sin 2a
(D) 1 0sin 4a
In the circumference with radius 50 cm is inscribed a quadrilateral. Two of its angles are 450 and 1200. The length of diagonals is : (B) 10 2 cm; 10 3 cm (A) 25 2 cm; 25 3 cm (C) 50 2 cm; 50 3 cm
(D) None of these
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12.
In ABC, A = 450, B = 300. M is a point on the side AB. The radius of the circumcircle of AMC is R. The radius of the circumcircle of MBC is : (A) 2R cm
13. 14. 15. 16. 17.
(B) R 2 cm
(A) 5 : 1 (B) ( 5 1) : 1 (C) 1 : ( 5 1) (D) None of these In a ABC, if a sin A = b sin B, then the triangle is : (A) Right angled (B) Equilateral (C) Right angled isosceles (D) Isosceles Points D, E are taken on the side BC of a triangle ABC such that BD = DE = EC. If BAD = x, DAE = y,
sin( x y ) sin( y z ) is equal to : sin x sin z
(B) 2
(C) 4b
(D) None of these
In a triangle ABC, A = 45 , B = 75 , then a 2 c is equal to : 0
(A) 2b
20.
(D) None of these
(A) 2 + 3 : 1 (B) 2 + 3 : 2 – 3 (C) 3 – 1 : 3 + 1 (D) None of these The perimeter of an acute angled triangle ABC is 6 times the arithmetic mean of the sines of its angles. If the side b is 2, the angle B is : (A) 300 (B) 600 (C) 900 (D) None of these If the angles of a triangle be in the ratio 1 : 4 : 5, then the ratio of it’s greatest side to the smallest side is :
(A) 1
19.
R cm 2
The angle of a triangle are as 5 : 5 : 2, the ratio of the greatest side to the least side is :
EAC = z, then the value of 18.
(C)
0
(B) b
(C) 4b
(D)
b 2
A hiker starts her journey at point A. She notices a farm house at point C and works out it’s bearing is at 1380. She then walks for 5 kilometres and stops at point B. At point B the hiker looks again at the farm house and calculates its bearing now to be 2000. The distance AC and BC respectively are :
(A) 3.28 km, 6.55 km (B) 2.66 km, 5.83 km (C) 2.83 km, 5.66 km (D) None of these The angles of a triangle are in the ratio 4 : 1 : 1, then the ratio of the largest side to the perimeter is (Use : sin (180 0 ) sin ) (A) 1 : (1 3 )
(B) 2 : 3
(C)
3 : (2 3 )
(D) 1 : (2 3 )
Based On Cosine Rule (Q. No. 21-37) 21.
In ABC, AB = 5 cm, AC = 6 cm, A = 600. The length of the side BC is :
22.
(A) 31 cm (B) 29 cm (C) 31 cm Which of the following options contains the sides of a right angled triangle ? (A) 13, 14, 15 (B) 12, 35, 37 (C) 13, 15, 24
(D) None of these
The size of C of ABC, if a = 2 3 cm, b = 3 cm, c = (A) 900 (B) 600 (C) 300
(D) None of these
23.
(D) 29 cm
3 cm is : 45
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24.
The size of C of ABC, if a = 11 cm, b = 60 cm, c = 61 cm is : (A) 900 (B) 600 (C) 300
25.
In ABC we have AC = 3 cm, BC =
26.
(A) 3 cm (B) 3 3 cm The length of a diagonal of a rectangle is 32 cm, and the angle between the diagonals is 1350. The length of the sides of rectangle are :
5 cm, A = 45 . The length of the side AB is : (C) 2 cm or 2 2 cm (D) 3 cm or 3 3 cm
(A) 4 3 3 cm and 4 3 3 cm (C) 4 cm and 16 cm 27.
28.
29.
(D) None of these
0
(B) 16 2 2 cm and 16 2 2 cm (D) None of these
The incentre of a right angled triangle is at distance 5 and length of the hypotenuse is : (A) 5 cm (B) 10 cm (C) 15 cm
10 from the two ends of the hypotenuse. The (D) 7.5 cm
The incentre of ABC is at distance 7 and 3 3 from the point A and B. If the angle at point C is 1200, the length of the side AB is : (B) 129 cm (C) 119 cm (D) None of these (A) 139 cm Calculate the length y of the side in the triangle below :
A y
5 420
B
30.
31. 32.
33.
C
(A) 5.25 (B) 4 (C) 6.25 (D) None of these A ship sails from harbour and travels 25 km on a bearing of 300 before reaching a marker bouy. At this point the ship turns and follows a course on a bearing of 900 and travels for 32 km until it reaches an island. On the return journey, the ship is able to take the most direct route back to the harbour. The total distance triavelled by the ship is : (A) 105 km (B) 95 km (C) 112 km (D) 130 km If the angles of a triangle ABC are in AP, then : (A) c2 = a2 + b2 + ab (B) a2 + c2 – ac = b2 2 2 2 (C) c = a + b (D) None of these If a = 4, b = 3 and A = 600, then c is a root of the equation : (A) x2 – 3x – 7 (B) x2 + 3x + 7 = 0 2 (C) x – 3x + 7 (D) x2 + 3x – 7 = 0 If p1, p2, p3 are the altitudes of a triangle from the vertices A, B C and , the area of the triangle, =
34.
9
1 1 1 p1 p2 p3
ab(1 k ) , then k is equal to : (a b c)
(A) cos C
(B) cos A
(C) cos B
(D) None of these
(A) a2 + b2 – c2
(B) c2 + a2 – b2
(C) b2 – c2 – a2
(D) c2 – a2 – b2
A B C In a ABC, 2ac sin is equal to : 2
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35.
In a triangle the length of two larger sides are 10 and 9 respectively. If the angles are in A. P., then the third side can be : [DCE-2001] (A) 5 6
36.
(B) 5 6
In a ABC if b = 20, c = 21 and sin A = (A) 12
37.
(C) 3 3
In a ABC, (A)
5 7
3 , then a = 5
(B) 13
(D) 5 [EAMCET-2003]
(C) 14
bc ca ab , then cos C = 11 12 13 7 16 (B) (C) 5 17
(D) 15 [Karnataka-CET-2003] (D)
17 36
Mixed Applications of Sine & Cosine Rule (Q.No. 38-41) 38.
The sides of a triangle are 3 1 and is : (A) 300 (B) 450
3 1 and the included angle is 600. The difference of the remaining angles (C) 600
(D) 900
39.
If two sides of a triangle and the included angle are given by a = (1 3 ) cm, b = 2 cm, c = 600, the other two
40.
angles are : (A) 900, 300 (B) 750, 450 In the previous Q., the third side is ;
41.
(B) 6 cm (A) 6 cm 2 2 2 If b + c = 3a , then cot B + cot C – cot A = (A) 1
(B)
ab 4
(C) 600, 600
(D) None of these
(C) 9 cm
(D) None of these
(C) 0
(D)
ac 4
Based On Area Of Triangle (Q.No. 42-46) 42.
In a triangle ABC, B = 450, a = 2( 3 1 ) and area of ABC = 6 + 2 3 square units, then the side b is equal to (A)
3 1 2
(B) 4
(C)
2 ( 3 1)
(D) None of these
(a b c)(c b a)(c a b)(a b c) is equal to : 4b 2 c 2
43.
In any ABC, the expression
44.
(A) cos2A (B) sin2A (C) 1 – cosA (D) 1 + cos A In any ABC, the expression (a + b +c) (a + b – c) (b + c – a) (c + a – b) is equal to : (A) 16
45.
47.
(C) 4
(D) None of these
bx cy az = c a b 2(a 2 b 2 c 2 ) (D) R
If x,y,z are perpendiculars drawn from the vertices of a triangle having sides a, b and c, then
a2 b2 c2 a2 b2 c2 (C) R 4R In an equilateral triangle of each side 2 3 cm, the radius of the circum circle is :
(A) 46.
(B) 4 2
a2 b2 c2 2R
(B)
(A) 2 cm (B) 1 cm (C) 3 cm (D) 2 3 cm A pole stands vertically inside a triangular partk ABC. If the angle of elevation of the top of the pole from each [IIT-2001] corner of the park is same, then in ABC, the foot of the pole is at the : (A) Centroid (B) Circumcentre (C) Incentre (D) Orthocentre
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48.
A man from the top of a 100 m high tower sees a car moving towards the tower at an angle of depression of 300. After some time, the angle of depression becomes 600. the distance (in metres) traveled by the car during this time is :
[II’T-Screening-2001]
3
(A) 100 49.
(B)
(C)
100 3 3
(D) 200 3
The value of k for which (cosx + sinx)2 + k in sin x cos x – 1 = 0 is an identity is : [Kerala Engineering-2001] (A) – 1
50.
200 3 3
(B) – 2
(C) 0
(D) 1
Which of the following pieces of does not uniquely determine an acute angled triangle ABC (R beign the radius of the circumcircle)? (A) a, sinA, sinB
(B) a, b, c
51.
The value of
(A) a, sinA, sinB 52.
cos 2 (A)
53.
12
(C) a, sinB, R
(D) a, sinA, R.
1 tan 15 1 tan 2 150 2
cos 2
4
0
[II’ Screening-2002]
(B) a, b, c
cos 2
2 3 3
(C) a, sinB, R
(D) a, sinA, R.
5 is equal to : 12
(B)
2 3
[Karnataka-CET-2002] (C)
3 3 2
(D)
If tanA + cotA = 4, then tan4A + cot4A is equal to :
[Kerala Engineering-2002]
(A) 110
(D) 194
(B) 191
(C) 80
54.
If tan sec e , then cos equals :
55.
2 e x ex e x e x (B) x (C) e ex 2 2 2 In a ABC, if a2 + b2 + c2 – ab – bc – ca = 0, then sin A + sin2B + sin2C =
x
[AMU-2002]
(A)
(A) 56.
4 9
(B)
9 4
(C) 3 3
In a triangle ABC, medians AD and BE are drawn. If AD = 4, DAB
(A)
64 3
(D)
e x e x e x ex
[Karnataka-CET-2003] (D) 1
triangle ABC is :
57.
2 3
6
and ABE
3
, then the area of the
[AIEEE-2003] (B)
8 3
(C)
32 3
(D)
32 3 3
3 5
3 4
The upper the portion of a vertical pole subtends an angle tan–1 at a point in the horizontal plane through it’s foot and at a distance 40 m from the foot. A possible height of the vertical pole is : [Hint : Use the formula tan( ) (A) 60 m
(B) 20 m
tan tan ] 1 tan tan
(C) 40 m
[AIEEE-2003] (D) 80 m
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58.
1 1 . cos , then the value of lies in : [IIT-Screening-2004] 2 3 2 2 5 5 , (B) , (C) (D) , 3 3 3 6 6
If and are acute angles, sin
, 3 2
(A) 59.
The sides of a triangle are in the ratio 1 :
3 : 2, the angles of the triangle are in the ratio :
(A) 1 : 3 : 5
(B) 2 : 3 : 4
(C) 3 : 2 : 1
(D) 1 : 2 : 3
[IIT-Screening-2004]
60.
A person standing on the bank of a river observes that the angle of elevation of the top of a tree on the opposite bank of the river is 600 and when he retires 40 metres away from the tree the angle of elevation becomes 300. The breadth of the river is : [AIEEE-2004] (A) 20 m (B) 30 m (C) 40 m (D) 60 m
61.
If the roots of the quadratic equation x2 + px + q = 0 are tan 300 and tan 150, then the value of 2 + q – p is (A) 1 (B) 2 [AIEEE-2006] (C) 3 (D) 0
62.
A tower stands at the centre of a circular park. A and B are two points on the boundary of the park such that AB (= a) subtends an angle of 600 at foot of the tower, and the angle of elevation of the top of the tower from A or B is 300. The height of the tower is : [AIEEE-2007]
2a 3
(A) 63.
(B) 2a 3
a 3
(C)
(D) a 3
AB is a vertical pole with B at the ground level and A at the top. A man finds that the angles of elevation of the point A from a certain point C on the ground is 600. He moves away from the pole along the line BC to a point D such that CD = 7 m. From D the angle of elevation of the point A is 450. then the height of the pole is : (A)
7 3 m 2( 3 1)
(B)
7 3 ( 3 1) m 2
(C)
7 3 ( 3 1) m 2
(D)
7 3 m 2( 3 1)
OBJECTIVE Que. Ans. Que. Ans. Que. Ans. Que. Ans. Que. Ans.
1 A 16 D 31 B 46 A 61 C
2 B 17 A 32 A 47 B 62 C
[AIEEE-2008]
EXERCISE -5
ANSWER KEY 3 A 18 A 33 A 48 B 63 B
4 B 19 C 34 B 49 B
5 A 20 C 35 A 50 D
6 A 21 A 36 B 51 C
7 C 22 B 37 A 52 D
8 A 23 C 38 D 53 D
9 C 24 A 39 B 54 B
10 B 25 C 40 B 55 B
11 C 26 B 41 C 56 D
12 B 27 A 42 B 57 C
13 A 28 A 43 B 58 B
14 C 29 C 44 D 59 D
15 B 30 B 45 A 60 A
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6’ 00.1
12, 00.2
18’ 00.3
24’ 00.4
Sin x0 30, 36’ 00.5 00.6
0000 0175 0349 0523 0698 0872 1045 1219 1392 1564 1736 1908 2079 2250 2419 2588 2756 2924 3090 3256 3420 3584 3746 3907 4067 4226 4384 4540 4695 4848 5000 5150 5299 5446 5592 5736 5878 6018 6157 6293 6428 6561 6691 6820 6947 7071
0017 0192 0366 0541 0715 0889 1063 1236 1409 1582 1754 1925 2096 2267 2436 2605 2773 2940 3107 3272 3437 3600 3762 3923 4083 4242 4399 4555 4710 4863 5015 5165 5314 5461 5606 5750 5892 6032 6170 6307 6441 6574 6704 6833 6959 7083
0035 0209 0384 0558 0732 0906 1080 1253 1426 1599 1771 1942 2113 2284 2453 2622 2790 2957 3123 3289 3453 3616 3778 3939 4099 4258 4415 4571 4726 4879 5030 5080 5329 5476 5621 5764 5906 6046 6184 6320 6455 6587 6717 6845 6972 7096
0052 0227 0401 0576 0750 0924 1097 1271 1444 1616 1788 1959 2300 2130 2470 2639 2807 2974 3140 3305 3469 3633 3795 3955 4115 4274 4431 4586 4741 4894 5045 5195 5344 5490 5635 5779 5920 6060 6198 6334 6468 6600 6730 6858 6984 7108
0070 0244 0419 0593 0767 0941 1115 1288 1461 1633 1805 1977 2147 2317 2487 2656 2823 2990 3156 3322 3486 3649 3811 3971 4131 4289 4436 4602 4756 4909 5060 5210 5358 5505 5650 5793 5934 6074 6211 6347 6481 6613 6743 6871 6997 7120
0087 0262 0436 0610 0785 0958 1132 1305 1478 1650 1822 1994 2164 2334 2504 2672 2840 3007 3173 3338 3502 3665 3827 3987 4147 4305 4462 4617 4772 4924 5075 5225 5373 5519 5664 5807 5948 6088 6225 6361 6494 6626 6756 6884 7009 7133
Degr ee
P
.0 .1 .2 .3 .4 .5 .6 .7 .8 .9 .10 .11 .12 .13 .14 .15 .16 .17 .18 .19 .20 .21 .22 .23 .24 .25 .26 .27 .28 .29 .30 .31 .32 .33 .34 .35 .36 .37 .38 .39 .40 .41 .42 .43 .44 .45
0105 0279 0454 0628 0802 0976 1149 1323 1495 1668 1840 2011 2181 2351 2521 2689 2857 3024 3190 3355 3518 3681 3843 4003 4163 4321 4478 4433 4787 4939 5090 5240 5388 5534 5678 5821 5962 6101 6239 6374 6508 6639 6769 6896 7022 7145
42’ 00.7
48, 00.8
54’ 00.9
0122 0297 0471 1645 1819 1993 1167 1340 1513 1685 1857 2028 2198 2368 2538 2706 2874 3040 3206 3371 3535 3697 3859 4019 4179 4337 4493 4648 4802 4955 5105 5255 5402 5548 5693 5835 5976 6115 6252 6388 6521 6652 6782 6909 7034 7157
0140 0314 0488 0663 2837 1011 1184 1357 1530 1702 1874 2045 2215 2385 2554 2723 2890 3057 3223 3387 3551 3714 3875 4035 4195 4352 4509 4664 4818 4970 5120 5270 5417 5563 5707 5850 5850 5990 6129 6266 6401 6534 6665 6794 6921 7046
0157 0332 0506 0680 0854 1028 1201 1374 1547 1719 1891 2062 2233 2402 2571 2740 2907 3074 3239 3404 3567 3730 3891 4051 4210 4368 4524 4679 4833 4985 5135 5284 5432 5577 5721 5864 6004 6143 6280 6414 6547 6678 6807 6934 7059 7181
MEAN DEFFERENCES 1 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
2 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 4 4 4 4 4 4 4
3 9 9 9 9 9 9 9 9 9 9 9 9 9 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 7 7 7 7 7 7 7 7 7 7 7 6 6 6 6
4 12 12 12 12 12 12 12 12 12 12 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 10 10 10 10 10 10 10 10 10 9 9 9 9 9 9 9 9 8 8 8
5 15 15 15 15 15 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 13 13 13 13 13 13 13 12 12 12 12 12 12 12 12 11 11 11 11 11 10 10
50
R E
-F O
U N D A T IO N
B Y
MANISH
MATHEMATICS
KUMAR 0’ 00.0
6’ 00.1
12, 00.2
18’ 00.3
24’ 00.4
Sin x0 30, 36’ 00.5 00.6
7193 7314 7431 7547 7660 7771 7880 7986 8090 8192 8290 8387 8480 8572 8660 8746 8829 8910 8988 9063 9135 9205 9272 9336 9397 9455 9511 9563 9613 9659 9403 9744 9781 9816 9848 9877 9903 9925 9945 9962 9976 9986 9994 9998
7206 7325 7443 7559 7672 7782 7891 7997 8100 8202 8300 8396 8490 8581 8669 8755 8838 8318 8996 9070 9193 9212 9278 9342 9403 9461 9516 9568 9617 9664 9707 9748 9785 9820 9851 9880 9905 9928 9947 9963 9977 9987 9995 9999
7218 7337 7455 7570 7683 7793 7902 8007 8111 8211 8310 8406 8499 8590 8678 8763 8846 8926 9003 9078 9150 9219 9285 9348 9409 9466 9521 9573 9622 9668 9711 9751 9789 9823 9854 9882 9907 9930 9949 9965 9978 9988 9995 9999
7230 7349 7466 7581 7694 7804 7912 8018 8121 8221 8320 8415 8508 8599 8686 8771 8854 8934 9011 9085 9157 9225 9291 9354 9415 9472 9527 9578 9627 9673 9715 9755 9792 9826 9857 9885 9910 9932 9951 9966 9979 9989 9996 9999
7242 7361 7478 7593 7705 7815 7923 8028 8131 8231 8329 8425 8517 8607 8695 8780 8862 8942 9018 9092 9194 9232 9298 9361 9421 9478 9532 9583 9632 9677 9720 9759 9796 9829 9860 9888 9912 9934 9952 9968 9980 9990 9996 9999
7256 7373 7490 7604 7716 7826 7934 8039 8141 8241 8339 8434 8526 8616 8704 8788 8870 8949 9026 9100 9171 9239 9304 9367 9426 9483 9537 9588 9636 9681 9724 9763 9799 9833 9863 9890 9914 9936 9954 9969 9981 9990 9997 1.000
Degre e
P
.46 .47 .48 .49 .50 .51 .52 .53 .54 .55 .56 .57 .58 .59 .60 .61 .62 .63 .64 .65 .66 .67 .68 .69 .70 .71 .72 .73 .74 .75 .76 .77 .78 .79 .80 .81 .82 .83 .84 .85 .86 .87 .88 .89
7266 7385 7501 7615 7727 7837 7944 8049 8151 8251 8348 8443 8536 8625 8712 8796 8878 8957 9033 9107 9178 9245 9311 9373 9432 9489 9542 9593 9641 9686 9728 9767 9803 9836 9866 9893 9917 9938 9956 9971 9982 9991 9997 1.000
42’ 00.7
48, 00.8
54’ 00.9
7278 7396 7513 7627 7738 7848 7955 8059 8161 8261 8358 8453 8545 8634 8721 8805 8886 8965 9041 9114 9184 9252 9317 9379 9438 9494 9548 9598 9646 9690 9732 9770 9806 9839 9869 9895 9919 9940 9957 9972 9983 9992 9997 1.000
7290 7408 7524 7638 7749 7859 7965 8070 8171 8271 8368 8462 8554 8643 8729 8813 8894 8973 9048 9121 9191 9259 9323 9385 9444 9500 9553 9603 9650 9694 9736 9774 9810 9842 9871 9898 9921 9942 9959 9973 9984 9993 9998 1.000
7302 7420 7536 7649 7760 7869 7976 8080 8181 8281 8377 8471 8563 8652 8738 8821 8902 8980 9056 9128 9198 9265 9330 9391 9449 9505 9558 9608 9655 9699 9740 9778 9813 9845 9874 9900 9923 9943 9960 9974 9985 9993 9998 1.000
MEAN DEFFERENCES 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0
2 4 4 4 4 4 4 4 3 3 3 3 3 3 3 3 3 3 3 3 2 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0
3 6 6 6 6 6 5 5 5 5 5 5 5 5 4 4 4 4 4 4 4 3 3 3 3 3 3 3 2 2 2 2 2 2 2 1 1 1 1 1 1 1 0 0 0
4 8 8 8 8 7 7 7 7 7 7 6 6 6 6 6 6 5 5 5 5 5 4 4 4 4 4 3 3 3 3 3 3 2 2 2 2 2 1 1 1 1 1 0 0
5 10 10 10 9 9 9 9 9 8 8 8 8 8 7 7 7 7 6 6 6 6 6 5 5 5 5 4 4 4 4 3 3 3 3 2 2 2 2 2 1 1 1 0 0
51
R E
-F O
U N D A T IO N
B Y
MANISH
MATHEMATICS
KUMAR
Degree
P
0’ 00.0
.0 .1 .2 .3 .4 .5 .6 .7 .8 .9 .10 .11 .12 .13 .14 .15 .16 .17 .18 .19 .20 .21 .22 .23 .24 .25 .26 .27 .28 .29 .30 .31 .32 .33 .34 .35 .36 .37 .38 .39 .40 .41 .42 .43 .44 .45
1.000 .9998 .9994 .9986 .9976 .9962 .9945 .9925 .9903 .9877 .9848 .9816 .9781 .9744 .9703 .9659 .9613 .9563 .9511 .9455 .9397 .9336 .9272 .9205 .9135 .9063 .8988 .8910 .8829 .8746 .8660 .8575 .8480 .8387 .8290 .8192 .8090 .7986 .7880 .7771 .7660 .7547 .7431 .7314 .7193 .7071
6’ 00.1 1.000 9998 9993 9985 9974 9960 9943 9923 9900 9874 9845 9813 9778 9740 9699 9655 9608 9558 9505 9449 9391 9330 9265 9198 9128 9056 8980 8902 8821 8738 8652 8563 8471 8377 8281 8181 8080 7976 7859 7760 7649 7536 7420 7302 7181 7059
12, 00.2 1.000 9998 9993 9984 9973 9959 9942 9921 9898 9871 9842 9810 9774 9736 9694 9650 9603 9553 9500 9444 9385 9323 5259 9191 9121 9048 8973 8894 8813 8729 8643 8554 8462 8368 8271 8171 8070 7965 7869 7749 7638 7424 7408 7290 7169 7046
18’ 00.3 1.000 9997 9992 9983 9972 9957 9940 9919 9895 9869 9839 9806 9770 9732 9690 9646 9598 9548 9494 9438 9379 9317 9252 9184 9114 9041 8965 8886 8805 8721 8634 8545 8453 8358 8261 8161 8059 7955 7848 7738 7627 7513 7396 7278 7157 7034
24’ 00.4 1.000 9997 9991 9982 9971 9956 9938 9917 9893 9866 9836 9805 9767 9728 9686 9641 9593 9542 9489 9432 9373 9311 9245 9178 9107 9033 8957 8878 8796 8712 8625 8536 8443 8343 8251 8151 8049 7944 7837 7727 7615 7501 7385 7266 7145 7022
30, 00.5 1.000 9997 9990 9981 9969 9954 9936 9914 9890 9863 9833 9799 9763 9724 9681 9636 9588 9537 9583 9426 9367 9304 9239 9171 9100 9026 8949 8870 8788 8704 8616 8526 8434 8339 8241 8141 8039 7934 7826 7716 7604 7490 7373 7254 7133 7009
36’ 00.6 9999 9996 9990 9980 9968 9952 9934 9912 9888 9860 9829 9796 9759 9720 9677 9632 9583 9532 9478 9421 9361 9298 9232 9164 9092 9018 8942 8862 8780 8695 8607 8517 8425 8329 8231 8131 8028 7923 7815 7705 7593 7478 7361 7242 7120 6997
42’ 00.7 9999 9996 9989 9966 9979 9951 9932 9910 9885 9857 9826 9792 9755 9715 9673 9627 9578 9527 9472 9415 9354 9291 9225 9157 9085 9011 8934 8854 8771 8686 8599 8508 8415 8320 8221 8121 8018 7912 7804 7694 7581 7466 7349 7230 7108 6984
48, 00.8 9999 9995 9988 9978 9965 9949 9930 9907 9882 9854 9823 9789 9751 9711 9668 9622 9573 9521 9466 9409 9348 9285 9219 9150 9078 9003 8926 8846 8763 8678 8590 8499 8406 8310 8211 8111 8007 7902 7793 7683 7570 7455 7337 7218 7096 6972
54’ 00.9 9999 9995 9987 9977 9963 9947 9928 9905 9880 9851 9820 9785 9748 9707 9664 9617 9568 9516 9461 9403 9342 9278 9212 9143 9070 8996 8918 8838 8755 8669 8581 8490 8396 8300 8202 8100 7997 7891 7782 7672 7559 7443 7325 7206 7083 6959
MEAN EFFERENCES 1 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
2 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 3 3 4 4 4 4 4 4 4 4 4
3 0 0 0 1 1 1 1 1 1 1 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 4 4 4 4 4 4 5 5 5 5 5 5 5 5 6 6 6 6 6 6 6
4 0 0 1 1 1 1 1 2 2 2 2 2 3 3 3 3 3 3 4 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 7 7 7 7 7 7 8 8 8 8 8 8
52
5 0 0 1 1 1 2 2 2 2 2 3 3 3 3 4 4 4 4 5 5 5 5 6 6 6 6 6 7 7 7 7 8 8 8 8 8 9 9 9 9 9 10 10 10 10 10
R E
-F O
U N D A T IO N
B Y
MANISH
MATHEMATICS
KUMAR
Degree
P
0’ 00.0
.46 .47 .48 .49 .50 .51 .52 .53 .54 .55 .56 .57 .58 .59 .60 .61 .62 .63 .64 .65 .66 .67 .68 .69 .70 .71 .72 .73 .74 .75 .76 .77 .78 .79 .80 .81 .82 .83 .84 .85 .86 .87 .88 .89
.6947 .6820 .6691 .6428 .6428 .9293 .6157 .6018 .5878 .5736 .5592 .5446 .5299 .5150 .5000 .4848 .4695 .4540 .4384 .4226 .4067 .3907 .3746 .3584 .3420 .3256 .3090 .2924 .2756 .2588 .2419 .2250 .2079 .1908 .1736 .1564 .1392 .1219 .1045 .0872 .0698 .0523 .0349 .0175
6’ 00.1 6934 6807 6678 6547 6414 6280 6124 6004 5864 5721 5577 5432 5284 5135 4985 4833 4679 4524 4368 4210 4051 3891 3730 3567 3404 3239 3074 2907 2740 2571 2402 2233 2062 1891 1719 1547 1374 1201 1028 0854 0680 0506 0332 0157
12, 00.2 6921 6794 6665 6534 6401 6266 6129 5990 5850 5707 5563 5417 5270 5120 4970 4818 4664 4509 4352 4195 4035 3875 3714 3551 3387 3223 3057 2890 2723 2554 2385 2215 2045 1874 1702 1530 1357 1184 1011 0837 0663 0488 0314 0140
18’ 00.3 6909 6782 6652 6521 3688 6252 6115 5976 5835 5693 5548 5402 5255 5105 4955 4802 4648 4493 4337 4179 4019 3859 3697 3535 3371 3206 3040 2874 2706 2538 2368 2198 2028 1857 1685 1513 1340 1167 0993 0819 0645 0471 0297 0122
24’ 00.4 6896 6769 6639 6508 6374 6239 6101 5962 5821 5678 5534 5388 5240 5090 4939 4787 4633 4478 4321 4163 4003 3843 3681 3518 3335 3190 3024 2857 2689 2521 2351 2181 2011 1840 1668 1495 1323 1149 0976 0802 0623 0454 0279 0105
30, 00.5 6884 6756 6626 6494 6361 6225 6088 5948 5807 5664 5519 5373 5225 5075 4924 4772 4617 4462 4305 4147 3987 3887 3665 3502 3338 3173 3007 2840 2672 2504 2334 2164 1994 1822 1650 1478 1305 1132 0958 0785 0610 0436 0262 0087
36’ 00.6 6871 6743 6613 6481 6347 6211 6074 5934 5793 5650 5505 5358 5210 5060 4909 4756 4602 4446 4289 4131 3971 3811 3649 3486 3322 3156 2990 2823 2656 2487 2317 2147 1977 1805 1633 1461 1288 1115 0941 0767 0593 0419 0244 0070
42’ 00.7 6858 6730 6600 6468 6334 6198 6060 5920 5779 5635 5490 5344 5135 5045 4894 4741 4586 4431 4274 4115 3955 3795 3633 3469 3305 3140 2974 2807 2639 2470 2300 2130 1959 1788 1616 1444 1271 1097 0924 0750 0576 0401 0227 0052
48, 00.8 6845 6717 6587 6455 6320 6184 6064 5906 5764 5621 5476 5329 5180 5030 4879 4726 4571 4415 4258 4099 3939 3778 3616 3453 3289 3123 2957 2790 2622 2453 2284 2113 1942 1771 1599 1426 1253 1080 0906 0732 0558 0384 0209 0035
54’ 00.9 6833 6704 6574 6441 6307 6170 6032 5892 5750 5606 5461 5314 5165 5015 4863 4710 4555 4399 4242 4083 3923 3762 3600 3437 3272 3107 2940 2773 2605 2436 2267 2096 1925 1754 1582 1409 1236 1063 0889 0715 0541 0366 0192 0017
MEAN EFFERENCES 1 2 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3
2 4 4 4 4 4 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6
3 6 6 7 7 7 7 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 9 9 9 9
4 8 9 9 9 9 9 9 9 9 10 10 10 10 10 10 10 10 10 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 12 12 12 12 12 12 12 12 12 12
53
5 11 11 11 11 11 11 12 12 12 12 12 12 12 12 13 13 13 13 13 13 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 15 15 15 15
R E
-F O
U N D A T IO N
B Y
MANISH
MATHEMATICS
KUMAR 0’ 00.0
6’ 00.1
12, 00.2
18’ 00.3
24’ 00.4
tan x0 30, 36’ 00.5 00.6
0000 0175 0349 0524 0699 0875 1051 1228 1405 1584 1763 1944 2126 2309 2493 2679 2867 3057 3249 3443 3640 3839 4040 4245 4452 4663 4877 5095 5317 5543 5774 6009 6249 6494 6745 7002 7265 7536 7513 8098 8391 8693 9004 9325 9657 1.000
0017 0192 0367 0542 0717 0892 1069 1246 1423 1602 1781 1962 2144 2327 2512 2698 2886 3076 3269 3463 3659 3859 4061 4265 4473 4684 4809 5117 5340 5566 5797 6032 6273 6519 6771 7028 7292 7563 7841 8127 8421 8724 9036 9358 9691 0035
0035 0209 0384 0559 0734 0910 1086 1263 1441 1620 1799 1980 2162 2345 2530 2717 1905 3096 3288 3482 3679 3869 4081 4286 4494 4706 4921 5139 5362 5589 5820 6056 6297 6544 6796 7054 7319 7590 7869 8156 8451 8754 9067 9391 9725 0070
0052 0227 0402 0577 0752 0928 1104 1281 1459 1638 1817 1998 2180 2364 2549 2736 2924 3115 3307 3502 3699 3899 4101 4307 4515 4727 4942 5161 5384 5612 5844 6080 6322 6569 6822 7080 7346 7618 7898 8185 8481 8785 9099 9424 9759 0105
0070 0244 0419 0594 0769 0945 1122 1299 1477 1635 1835 2016 2199 2382 2568 2754 2943 3134 3327 3522 3719 3919 4122 4327 4536 4748 4962 5184 5407 5635 5867 6104 6346 6594 6874 7107 7373 7646 7926 8214 8511 8816 9131 9457 9793 0141
0087 0262 0437 0612 0787 0963 1139 1317 1495 1673 1853 2035 2217 2401 2586 2773 2962 3253 3541 3739 4142 4348 4557 4770 4986 5206 5430 5658 5890 6128 6371 6619 6873 7133 7400 7673 7400 7673 7954 8243 8541 8847 9163 9490 9827 0176
Degr ee
P
.0 .1 .2 .3 .4 .5 .6 .7 .8 .9 .10 .11 .12 .13 .14 .15 .16 .17 .18 .19 .20 .21 .22 .23 .24 .25 .26 .27 .28 .29 .30 .31 .32 .33 .34 .35 .36 .37 .38 .39 .40 .41 .42 .43 .44 .45
0105 0279 0454 0629 0805 0981 1157 1334 1512 1691 1871 2053 2235 2419 2605 2792 2981 3172 3365 3561 3759 3959 4163 4369 4578 4791 5008 5228 5452 5681 5914 6152 6395 6644 6899 7159 7427 7701 7983 8273 8571 8878 9195 9523 9861 0212
42’ 00.7
48, 54’ 00.8 00.9
0122 0297 0472 0647 0822 0998 1175 1352 1530 1709 1890 2071 2254 2438 2623 2811 3000 3191 3385 3581 3779 3979 4183 4390 4599 4813 5029 5250 5475 5704 5938 6176 6420 6669 6924 7186 7454 7729 8012 8302 8601 8910 9228 9556 9896 0247
0140 0314 0489 0664 0840 1016 1192 1370 1548 1727 1908 2089 2272 2456 6242 2830 3019 3211 3404 3600 3799 4000 4204 4411 4621 4834 5051 5272 5498 5727 5961 6200 6445 6694 6950 7212 7481 7757 8040 8332 8632 8941 6260 9590 9930 0283
0157 0332 0507 0682 0857 1033 1210 1388 1566 1745 1926 2107 2290 2475 2661 2849 3038 3230 3424 3620 3819 4020 4224 4431 4642 4856 5073 5295 5520 5750 5985 6224 6469 6720 6976 7239 7508 7785 8069 8361 8662 8972 9293 9623 9965 0319
MEAN DEFFERENCES 1 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 4 4 4 4 4 4 4 4 4 4 4 4 5 5 5 5 5 5 5 6 6 6
2 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 7 7 7 7 7 7 7 7 7 8 8 8 8 8 8 9 9 9 9 9 10 10 10 11 11 11 12
3 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 10 10 10 10 10 10 10 11 11 11 11 11 12 12 12 12 13 13 13 14 14 14 15 15 16 16 17 17 18
4 12 12 12 12 12 12 12 12 12 12 12 12 12 12 12 13 13 13 13 13 13 13 14 14 14 14 15 15 15 15 16 16 16 17 17 18 18 18 19 20 20 21 21 22 23 24
5 15 15 15 15 15 15 15 15 15 15 15 15 15 15 16 16 16 16 16 16 17 17 17 17 18 18 18 18 19 19 20 20 20 21 21 22 23 23 24 24 25 26 27 28 29 30
54
P
R E
-F O
U N D A T IO N
B Y
MANISH
MATHEMATICS
Degree
KUMAR
0’ 00.0
6’ 00.1
12, 00.2
18’ 00.3
24’ 00.4
30, 00.5
.46 .47 .48 .49 .50 .51 .52 .53 .54 .55 .56 .57 .58 .59 .60 .61 .62 .63 .64 .65 .66 .67 .68 .69 .70 .71 .72 .73 .74 .75 .76 .77 .78 .79 .80 .81 .82 .83 .84 .85 .86 .87 .88 .89
1.0355 1.0724 1.1106 1.1504 1.1918 1.2349 1.2799 1.3270 1.3764 1.4281 1.4826 1.5399 1.6003 1.6643 1.7321 1.8040 1.8807 1.9626 2.0503 2.1445 2.2460 2.3559 2.4751 2.6051 2.7475 2.9042 3.0777 .02709 3.4874 3.7321 4.0108 4.3315 4.7046 5.1446 5.6713 6.3138 7.1154 8.1443 9.5144 11.43 14.30 19.08 28.64 57.29
0392 0761 1145 1544 1960 2393 2846 3319 3814 4335 4882 5458 6066 6709 7391 8115 8887 9711 0594 1543 2566 3673 4876 6187 7625 9208 0961 2914 5105 7583 0408 3662 7453 1923 7297 3859 2066 2636 9.677 11.66 14.67 19.74 30.14 63.66
0428 0299 1184 1585 2002 2437 2892 3367 3865 4388 4938 4938 5517 6128 6775 7461 8190 8967 9797 0686 1642 2673 3789 5002 6325 7776 9375 1146 3122 5339 7848 0713 4015 7867 2422 7894 4596 3002 9.845 11.91 15.06 20.45 31.82 71.62
0464 0837 1224 1626 2045 2484 2938 3416 3916 4442 4994 5577 6191 6842 7532 8265 9047 9883 0778 1742 2781 3906 5129 6464 7929 9544 1334 3332 5576 8118 1022 4374 8288 2924 8502 5350 3962 5126 10.02 12.16 15.46 21.20 33.69 81.85
0501 0875 1263 1667 2088 2527 2985 3465 3968 4496 5051 5637 6255 6909 7603 8341 9128 9970 0872 1842 2889 4023 5257 6605 8083 9714 1524 3544 5816 8391 1335 4737 8716 3435 9124 6122 4947 6427 10.20 12.43 15.89 22.02 35.80 95.49
0838 0913 1303 1708 2131 2572 3032 3514 4019 4550 5108 5697 6319 6977 7675 8418 9210 2.0057 0965 1943 2998 4142 5386 6746 8239 9887 1716 3759 6059 8667 1653 5107 9152 3955 9758 6912 5958 7769 10.39 12.71 16.35 22.90 38.19 114.6
tan x0 36’ 42’ 00.6 00.7 0575 0951 1343 1750 2174 2617 3079 3564 4071 4605 5166 5757 6383 7045 7747 8405 9262 2.0145 1060 2045 3109 4262 5517 6889 8397 3.0061 1910 3977 6305 8947 1976 5483 9594 4486 6.0405 7720 6996 9152 10.58 13.00 16.83 23.86 40.62 143.2
0612 0990 1383 1792 2218 2662 3127 3613 4124 4659 5224 5818 6447 7113 7820 8572 9375 2.0233 1155 2148 3220 4383 5649 7034 8556 3.0237 2106 4197 6554 9232 2303 5864 5.0045 5026 6.1066 8548 8062 9.0579 10.78 13.30 17.34 24.90 44.07 191.0
48, 00.8 0649 1028 1423 1833 2261 2708 3175 3663 4176 4715 5282 5880 6512 7182 7893 8650 9458 2.0323 1155 2148 3220 4383 5649 7034 8556 3.0237 2106 4197 6554 9232 2303 5864 5.0045 5026 6.1066 8548 8062 9.0579 10.78 13.30 17.34 24.90 44.07 191.0
54’ 00.9 0686 1067 1463 1875 2305 2753 3222 3713 4229 4770 5340 5941 6577 7251 7966 8728 9542 2.0413 1348 2355 3445 4627 5916 7326 8878 3.0595 2506 4646 7062 9812 2972 6646 5.0970 6140 6.2432 7.0264 8.0285 9.3572 11.20 13.95 18.46 27.27 52.08 573.0
MEAN EFFERENCES 1 6 6 7 7 7 8 8 8 9 9 10 10 11 11 12 13 14 15 16 17 18 20 22 24 26 29 32 36 41 46 53
2 12 13 13 14 14 15 16 16 17 18 19 20 21 23 24 26 27 29 31 34 37 40 43 47 52 58 64 72 81 93 107
3 18 19 20 21 22 23 24 25 26 27 29 30 32 34 36 38 41 44 47 51 55 60 65 71 78 87 96 108 122 139 160
4 25 25 27 28 29 30 31 33 34 36 38 40 43 45 48 51 55 58 63 68 73 79 87 95 104 116 129 144 163 186 213
55
5 31 32 33 34 36 38 39 41 43 45 48 50 53 56 60 64 68 73 78 85 92 99 108 119 133 145 161 180 204 232 267
