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ACADEMY
TRIGONOMETRY
Trigonometrical Basics and Formulae
Relation between P, B, H:
H  P B , P  H B , B H P
2
2
2
2
2
H
P
2

Ist
sin   P / H
B
, , , A, B, C Names of angle between Base and Hypo
2nd
3rd
4th
sin   1/ cosec 
tan   sin  / cos 
sin  90     cos 
cos   B / H
cosec   1/ sin 
cot   cos  / sin 
tan   P / B
cos   1/ sec 
tan   cot   1
cosec   H / P
sec   1/ cos 
sec   H / B
tan   1/ cot 
cot   B / P
cot   1/ tan 
cos  90     sin 
sec  90     cosec 
cosec  90     sec 
tan  90     cot 
cot  90     tan 
5th
1.
sin   cos   1
(i)
sin2   1  cos2 
(iv)
sin   
(vi)
cos   
2.
sec2   tan 2   1
(i)
sec2   1  tan2 
(ii)
tan 2   sec2   1
(iv)
tan    sec2   1
(v)
tan   
3.
cosec 2  cot 2   1
(i)
cosec 2  1  cot 2 
(ii)
cot 2   cosec2  1
(iv)
cot    cosec 2  1
(v)
cot   
2
2
(ii)
cos2   1  sin2 
1  cos  1  cos  
1  sin  sec   1
Graphics By :- Roshan Dhawan
(iii)
sin    1  cos 2 
(v)
cos    1  sin 2 
(iii)
sec    1  tan 2 
 sec   1sec   1
(iii)
cosec    1  cot 2 
 cosec   1 cosec   1
-1-
Written By :- R. K. Badhan
ACADEMY
TRIGONOMETRY
TRIGONOMETRICAL TABLE

0°
30°
45°
sin 
0
1
2
1
2
60°
3
2
90°
1
Angle of Elevation
Angle of depression
cos
1
3
1
1
2
1
2
0
tan
0
1
3
1
cosec 
n. d.
2
sec
1
2
3
2
2
2
3
2
3
n. d.
1
n. d.
To look up w. r. t. horizontal line
To look down w. r. t. horizontal line
cot 
n. d.
3
1
1
3
0
Relation between radiuses, length of Chord, no. of sides, Central angle of a regular
polygon in a Circumcircle:

L
E
1.
L  2r sin  r 

2
2sin
2
O
D
180
L
F
2.
L  2r sin
r 
180

n
2sin
n
C
Relation between Degree and Minute:
B
L
1°
=
60`

90° =
89° 60`
Graphics By :- Roshan Dhawan
-2-
Written By :- R. K. Badhan
ACADEMY
1.
2.
3.
4.
5.
6.
7.
8.
9.
TRIGONOMETRY
CONCEPT : APPLICATION OF TRIGONOMETRIC TABLE
tan45
Find the value of:
sin 30  cos60
tan45
sec60 2sin90
Evaluate:


cosec 30 cot 45
cos0
1
Evaluate: cos2 30 cos2 45  4sec2 60  cos2 90  2tan2 60
2
1
1
Evaluate: sin2 30 cos2 45  4tan2 30  sin2 90  2cos 2 90 
2
24
4
3
Evaluate: cot 2 30  3sin 2 60  2cos ec 2 60  tan 2 30
3
4
sin 30  cos45  tan60
Evaluate:
cot 30  sin45  cos60
Evaluate:  cos0  sin45  sin30sin90  cos45  cos60 
Find the value of: 4  sin4 30  cos4 60   3  sin 2 45  2cos 2 45 
What should be subtracted from: 4  sin4 30  cos4 60   2  cos 2 45  sin 2 60  to get
0.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
cos2 A
For A = 30°, verify that:
 sin A  cosec A
sin A
For A = 30°, verify that: cos 3 A = 4cos 3 A – 3 cos A
For A  B  45 , verify that: sin(A + B) = sin A cos B + cos A sin B.
2tan A
If A = 30°, verify that: tan 2A =
1  tan 2 A
If A = 60° and B = 30°, verify that: cos(A – B) = cos A cos B + sin A sin B
2tan A
If A = 30°, verify that: sin2A =
1  tan 2 A
Verify that: sin 60° = 2 sin 30° cos 30°
For A = 60° and B = 30°, verify that: sin(A – B) = sin A cos B – cos A sin B
Using the formula: sin(A – B) = sin A cos B – cos A sin B, find the value of sin 15°.
Using the formula sin(A – B) = sin A cos B – cos A sin B, find the value of sin 30°.
Using the formula: sin(A + B) = sin A cos B + cos A sin B, find the values of sin 75°.
Graphics By :- Roshan Dhawan
-3-
Written By :- R. K. Badhan
ACADEMY
21.
Using the formula: tan 2 
tan30 
22.
23.
24.
25.
26.
27.
28.
29.
30.
31.
32.
33.
TRIGONOMETRY
2tan 
, find the value of tan 60° given that
1  tan 2 
1
.
3
Using the formula: sin  
1
1  cos 2
, find the value of sin 30° given that cos60 
2
2
CONCEPT: RELATION BETWEEN PYTHAGORAS THEORAM
AND TRIGONOMETRY
4sin   cos  5
If 3 tan  = 2, prove that:

2sin   cos  7
3
4sin   2cos 
If tan   , find the value of:
4sin   3cos 
4
5sin   3cos 
If 5tan   4, find the value of:
5sin   2cos 
3cos   2sin 
If 2tan   1, find the value of:
2cos   sin 
a
cos   sin 
If tan   , find the value of:
b
cos   sin 
4
2sin   3cos 
If tan   , find the value of:
3
2sin   3cos 
5
If tan A  , find the value of: sin A + cos A, where A < 90°.
12
7
If tan A   0  A  90  , find the value of: sin A + cos A.
24
(i)
If tan = 8/15 where   90 then find sin  .
2
(ii) If 5cos   6sin , find: 61  cos   sin  
1  sin  1
 .
1  sin  3
5sin A  3cos A
If 5 tan A = 4, find the value of:
4cos A  5sin A
4
If tan   , show that:
3
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-4-
Written By :- R. K. Badhan
ACADEMY
34.
If 3tan   4, find the value of:
35.
If sin  
36.
37.
38.
39.
40.
41.
42.
43.
44.
5sin   3cos 
.
5sin   2cos 
TRIGONOMETRY
5
and 0° < 0 < 90°, find the value of: cot  and cos  .
13
24
If sin A  , find the value of: tan A + sec A, where 0° < A < 90°.
25
8
If sin  
where  < 90° then find tan  .
17
3
If sin   , find the value of: tan   sec  .
5
3
If cos   , find the value of: cos   cosec  .
5
If a is an acute angle and tan a = 5/12, find the value of: cosec A.
If 3 tan   3sin  , find the value of sin2   cot 2 .
If tan   cot   2, find the value of: tan2   cot 2 .
If cot A = b/a, where a and b are real numbers, find the value of sin 2 A .
(i) Evaluate:  sec2   11  cosec 2   .
(ii) If sec   tan   p, prove that: sin  
p2  1
.
p2  1
CONCEPT : PROBLEMS BASED ON COMPLEMENTARY ANGLES
2
45.
46.
47.
48.
49.
2
 sin 35   cos55 
Find the value of: 
 
  2cos60
 cos55   sin 35 
cos70 cos59

 8sin 2 30
Evaluate:
sin 20 sin 31
cot 54 tan 20

2
Evaluate:
tan 36 cot 70
tan53 cot 80

Evaluate: 2
cot 37 tan10
tan 35 cot 78
Evaluate:

1
cot 55 tan12
Graphics By :- Roshan Dhawan
-5-
Written By :- R. K. Badhan
ACADEMY
TRIGONOMETRY
cot 50
sin 75
2 2
2
tan 40
cos 15
Find the value of:  sin72  cos18 sin72  cos18
Express cosec 69° + cot 69° in terms of angles between 0° and 45°.
Express cos 75° + cot 75° in terms of angles between 0° and 30°.
Find the value of sin2 35  sin2 55
2
2
 sin 27   cos63 
Evaluate: 
 

 cos63   sin 27 
Find the value of: cosec 2 67  tan2 23
Evaluate:
tan50  sec50
cos2 20  cos 2 70
(ii)
 cos40cosec 50
2
2
cot 40  cosec 40
sin 59  sin 31
sin15 cos75  cos15 sin 75
sec2 10  cot 2 80 
cos  sin  90     sin  cos  90   
cos58 sin 22
cos 38cosec 52


sin 32 cos68 tan18 tan 35 tan60 tan72 tan55
2sin68 2cot15 3tan45 tan 20 tan40 tan50 tan70


cos 22 5tan75
5
2
2
2
Evaluate: sin 20  sin 70  tan 45
Find the value of: sec 50° sin 40° + cos 40° cosec 50°
Find the value of: sec 70° sin 20° – cos 20° cosec 70°
Evaluate: tan 20° cot 70° – sec 20° cosec 70°
cos75 sin12
Evaluate:

 cos18cosec 72
sin15 cos78
Without using the table show that: sin 48° sec 42° + cos 48° cosec 42° = 2
Without using the table show that: sin 35° sin 55° – cos 35° cos 55° = 0
Without using the table, show that: tan 10° tan 15° tan 75° tan 80° = 1
Show that: sin 63° cos 27° + cos 63° sin 27° = 1
Show that: sin A cos (90° – A) + cos A sin (90° – A) = 1
sin 
cos 

 sec  cosec 
Show that:
sin  90    cos  90   
Evaluate: tan 35° tan 40° tan 45° tan 50° tan 55°
Evaluate: cosec  65     sec  25     tan  55     cot  35   
2
50.
51.
52.
53.
54.
55.
56.
57.
(i)
(iii)
(iv)
(v)
58.
59.
60.
61.
62.
63.
64.
65.
66.
67.
68.
69.
70.
2
Evaluate:
Graphics By :- Roshan Dhawan
-6-
Written By :- R. K. Badhan
ACADEMY
71.
72.
73.
Find the value of: sin  50     cos  40   
Find the value of: sin  55     cos  35   
Prove that:
(i)
tan10  tan75  tan15  tan80  1
(ii) tan1  tan2  tan3...........tan89  1
(iii) cos1  cos2  cos3..........cos180  0
(iv)
(v)
74.
TRIGONOMETRY
(i)
(ii)
sin2 5  sin2 10  ......  sin2 85  sin2 90  9
sin 3017
1
cos5943
If cos  81     sin  9    , find 
1
2
If tan2  cot    6 find .
If sin3  cos    6 , find  .
(i)
If A and B are the interior angles of a triangle ABC, right angled at C, prove
that: sin2 A  sin2 B  1 .
BC
(ii) If A, B, C are the interior angles of a triangle ABC, prove that: tan 
 =
 2 
A
cot   .
2
sin4 A  1
If sin A + cosec A = 3, find the value of:
sin 2 A
If A and B are acute angles and tan A = 1, sin B = 1/ 2, find the value of: cos (A +
B).
tan A  tanB
If tan A = ½ and tan B = 1/3, by using tan (A + B) =
, prove that: A +
1  tan AtanB
B = 45°.
tan 1  tan 2
Given that: tan  1  2  
where 1   2 are acute angles, calculate
1  tan 1 tan 2
1
1
1   2 where tan 1  and tan 2 
2
3
(iii)
75.
76.
77.
78.
79.
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-7-
Written By :- R. K. Badhan
ACADEMY
80.
(1)
TRIGONOMETRY
CONCEPT : TRIGONOMETRIC IDENTITIES
Prove the following identities:
1  sin2 A  sec2 A  1 (2) 1  tan2 A  cos2 A  1 (3)  cosec 2 A  1 tan2 A  1
(4)
1  cos A  cosec A  1 (5) 1  cot A  sin
(7)
cot 2  
(10)
(12)
(14)
2
2
sec A  1cosec A  1  1
1  tan   1  sin  1  sin  
2
2
2
2cos2  1
sin  cos 
sec2   cosec2   tan   cot  
(24)
1
 sec   tan 
sec   tan 
1  sin 
2
  sec   tan 
1  sin 
2
 sin   cos    1  2sin  cos  
(26)
(28)
tan2   cot 2   2  sec2  cosec2
1  cot   cosec 1  tan   sec   2
(20)
(22)
(29)
(30)
(32)
(33)
2
A 1
(6)
sec
2

A  1 cot 2 A  1
1
(8) sec A 1  sin 2 A  1
(9) cosec A 1  cos 2 A  1
 1
2
sin 
(11) 1  cos 2 A  sec 2 A  tan 2 A
1  cos A 1  cos A  1  cot 2 A   1
(16) cos   tan  
(18)
2
(13) secA 1  sin A secA  tan A   1
(15) secA 1  sin A sin A  tan A   1
(25)
1  tan2 
 tan2  where   45
2
cos   1
cos 
1  sin 

1  sin 
cos 
sec   1 1  cos 

sec   1 1  cos 
sec   tan 
 1  2sec  tan 2tan 2 
sec   tan 
sec2   cosec2  sec2  cosec2
(27)
tan2   sin2   tan2   sin2 
(17)
(19)
(21)
(23)
cosec   sin sec   cos  tan   cot    1
cosec A
cosec A
tan A
cot A
(31)

 2sec2 A

 sec A cosecA  1
cosec A  1 cosec A  1
1  cot A 1  tan A
2
2
 1  sin 2  
1  
1 

 tan   cos    tan   cos    2  1  sin 2  

 



1
1
1
1



 cosec  cot   sin  sin   cosec  cot  
Graphics By :- Roshan Dhawan
-8-
Written By :- R. K. Badhan
ACADEMY
(34)
(35)
(36)
(37)
(39)
TRIGONOMETRY
1
1
1
1



 sec   tan   cos  cos   sec   tan  
tan A  sec A  1 1  sin A

tan A  sec A  1
cos A
1
 cosec  sin   sec   cos   
tan   cot 
1 
1 
1
sin A  sinB cos A  cosB

(38)

0
 1  tan 2   1  cot 2    sin 2   sin4 
cos A  cosB sin A  sinB



sin8   cos8    sin 2   cos 2  1  2sin 2  cos 2  
sin2 A  sin2 B
(40) tan Atan B 
cos2 A  cos2 B
2
2
(41) 1  tan A tanB    tan A  tanB   sec 2 Asec 2 B
2
2
(42) sec6   tan6   3tan2  sec2   1
(43) tan2 A  cot 2 A  sec2 A  cosec2 A  2
(44)  secA  cosec A 1  tan A  cot A   tan AsecA  cot A cosecA
2
 1  sin   cos   1  cos 
(45) tan Asec B  sec Atan B  tan A  tan B
(46) 
  1  cos 
1

sin


cos



2
2
(47)  tan A  cosec B    cot B  sec A   2tan Acot B  cosec A  secB 
2
(48)
(49)
(50)
(52)
(53)
(55)
(56)
2
2
2
2
2
 sin A  sec A   cos A  cosec A   1  sec A  cosec A 
1  cot A  tan A sin A  cos A   sin2 A  cos2 A
2
2
sec3 A  cosec3 A
1  cos   sin  1  sin 

1  cos   sin 
cos 
(51)
2
1  cos   sin 2 
 cot 
sin  1  cos  
1
1
1  sin 2   cos 2 

 2
2
 sec2   cos 2  cosec2  sin 2   sin   cos   2  sin 2   cos 2 


sin A
cos A
1  sin  cot   cos 

 1 (54)

sec A  tan A  1 cosecA  cot A  1
1  sin  cot   cos 
sin4   cos4   sin 2   cos 2   2sin 2   1  1  2cos 2 

 sin
4
  cos4
 
   1  2sin
Graphics By :- Roshan Dhawan
2
 
 cos  
2
 
(57)  sec
-9-

4
 
  sec2   tan 4   tan 2 

Written By :- R. K. Badhan
ACADEMY
 sin
(59)
cos A
sin2 A

 sin Acos A
1  tan A cos A  sin A
(60)
2 sin   cos   3 sin   cos   1  0 (61)
(62)
(64)
(65)
(66)
(67)
(68)
81.
82.
83.
84.

 
TRIGONOMETRY
(58)
6
  cos   1  3sin  cos 
6
6
6
2
 
4
2
4


sin   2sin3 
(63)
 tan 
2cos3   cos 
sin   cos  sin   cos 
2


sin   cos  sin   cos   sin 2   cos 2  
1  sin A
1

 tan A
1  sin A cos A
tan3 
cot 3

 sec  cosec  2sin  cos 
1  tan2  1  cot 2 
sin2 Acos2 B  cos 2 Asin2 B  cos 2 Acos 2 B  sin 2 Asin 2 B  1
sin2 Acos2 B  cos2 Asin2 B  sin2 A  sin2 B
CONCEPT : PROVE USING GIVEN INFORMATIONS
If sec   tan   m and sec   tan   n , show that: mn = 1
(i)
If tan   sin   m and tan   sin   n show that: m2  n2  4 mn .
(ii) If sin   sin2   1 , prove that: cos2   cos4   1 .
a2 b2
If x  asin  and y  btan  , prove that: 2  2  1
x
y
(i)
If x  acos   bsin  and y  asin   bcos  , prove that: x 2  y 2   a 2  b 2  .
2
85.
86.
87.
cos3   sin3  cos3   sin3 

2
cos   sin 
cos   sin 
tan A  tanB
 tan AtanB
cot A  cot B
2
 x 3  y 3
If x  acos , y  bsin  , prove that:       1
a
b
x
y
x
y
If cos   sin   1 and sin   cos   1, prove that:
a
b
a
b
If l tan   m sec   n and l  tan   m sec   n show that:
2
2
 nl   ln 
 nm  mn 
 ml +lm   1   lm  l m 




3
Graphics By :- Roshan Dhawan
3
- 10 -
 x2 y2 
 2  2 2
b 
a
Written By :- R. K. Badhan
ACADEMY
88.
89.
TRIGONOMETRY
1
1
.
, then prove that: sec A + tan A = 2x or
4x
2x
cos 
cos 
If
 m and
 n , Show that:  m 2  n 2  cos 2   n 2 .
cos 
sin 
If sec A  x 
2
90.
91.
92.
2
 x h  y k 
If x  h  acos , y  k  bsin  . Prove that 
 
 1
 a   b 
If x  rsin  cos , y  rsin  sin , z  rcos  , show that: r 2  x 2  y 2  z 2
If x  acos   bsin  and y  asin   bcos  then show that: x 2  y 2  a2  b2
93.
If cosec   sin   l and sec   cos   m , prove that: l 2 m 2  l 2  m 2  3   1
94.
m2  1
If tan A = n tan B and sin A = m sin B, prove that: cos A  2
.
n 1
95.
2
If cosec   sin   m , and sec   cos   n , Prove that:  m 2 n 
2/ 3

 mn2

96.
If cosec   sin   a3 ,sec   cos   b3 , prove that : a 2b 2  a 2  b 2   1 .
97.
98.
If 2cos   sin   x and cos   3sin   y , prove that 2 x 2  y 2  2 xy  5
If a cos3   3a cos  sin3   m, a sin3   3a cos2  sin   n , prove that:
 m  n
2/ 3
  m  n
2/ 3
2/ 3
 1.
 2a 2 / 3
99. If a sin3 x  b cos3 x  sin x cos x and a sin x  b cos x, prove that : a 2  b2  1
100. (i) If (sec A + tan A) (sec B + tan B) (sec C + tan C) = (sec A – tan A). (sec B – tan B)
(sec C – tan C), prove that each is equal to  1.
(ii) If 7cosec   3cot   7 , prove that: 7cot   3 cosec   3 .
(iii) If a cos   b sin   c, prove that: a sin   b cos    a 2  b 2  c
101.
102.
103.
105.
CONCEPT : SOLVE TRIGONOMETRY EQUATIONS
4sin   3  0
If sin   cos   0,0    90, find the value of  .
1
2sin2   cos2   2
104. sin2   sin   0
2
106. 3tan   cot   5 cosec 
tan2   3  1 tan   3  0
2


Graphics By :- Roshan Dhawan
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Written By :- R. K. Badhan
ACADEMY
TRIGONOMETRY
cos 
3
cot   cos2 
cos 
cos 
109.

4
1  sin  1  sin 
2
107.
2
111. 2 cosec   3sec2 
108. sec2   2tan   0
cos 
cos 

2
cosec  1 cosec  1
cos 2   3cos   2
112.
1
sin2 
110.
CONCEPT : HEIGHTS AND DISTANCES
113. The length of the shadow of a pillar is 3 times its height. Find the angle of
elevation of the source of light.
114. The shadow of Qutab Minar is 81 m long when the angle of elevation of the sun is
41° 30`. Find the height of the Qutab Minar.
115. In the figure (1), ABCD is a rectangle with AD = 8 cm and CD = 12 cm. Line
segment CE is drawn making an angle of 60° with AB, intersecting AB in E. Find
the lengths of CE and BE upto 2 places of decimals.
116. In the figure (2), fine the length of AE.
117. In the figure (3), ABCD is a rectangle in which segment AP is drawn as shown. Find
the length of AP.
A
E
B
D
E
C
60°
40 cm
30°
90 cm
D
C
A
B
Fig 1
Fig 2
D
P
C
30°
30 cm
60 cm
A
B
Fig 3
118. The height of a pole is 20 m. It is broken by the wind in such a way that its top
touches the ground and makes an angle of 30° with the ground. Find the height
from the bottom to the point from where the pole was broken.
119. An aeroplane at an altitude of 200 m observes the angles of depression of opposite
points on the two banks of a river to be 45° and 60°. Find the width of the river.
Graphics By :- Roshan Dhawan
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Written By :- R. K. Badhan
ACADEMY
TRIGONOMETRY
120. From the top of a hill, the angles of depression of two consecutive kilometer stones
due east are found to be 30° and 45°. Find in meters, the height of the river.
121. The angle o elevation of the top of a tower at a point is 45° or 30°. After going 40 m
or 30 m towards the foot of the tower, the angle of elevation of the tower becomes
60°. Find the height of the tower.
122. The shadow of a tower, when the angle of elevation of sun is 45° is found to be 10
meters longer than when it was 60°. Find the height of the tower.
123. The angles of depression of two ships from the top of light house are 45° and 30°
towards east. If the ships are 200 m apart, find the height of the light house.
124. On the same side of a tower two objects are observed from the top of the tower,
their angles of depression are 45° and 60°. Find the distance between the objects if
the height of the tower is 300 m.
125. 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 60° when he moves 40 m away from
the bank he finds the angle of elevation to be 30°. Find the height of the tree and the
width of the river.
126. The length of the shadow of a tower standing on level ground is found to be 2x
meters longer when the Sun’s altitude is 30° than when it was 45°. Prove that the
height of tower is x 3  1 meters.


127. An aeroplane when 3000 m high passes vertically above another aeroplane at an
instance when their angles of elevation at the same observation point are 60° and
45° respectively. How many meters higher is one than the other?
128. From the top of a building 12 m high, the angle of elevation of the top of a tower is
found to be 45° and the angle of depression of the base of the tower at 30°. Find the
height of the tower and its distance on the ground from the building.
129. A pole 5 m high is fixed on the top of a tower. The angle of elevation of the top of the
pole observed from a point ‘A’ on the ground is 60° and the angles of depression of
the point ‘A’ from the top of the tower is 45°. Find the height of the tower.
130. A 7 m long flagstaff is fixed on the top of a tower on the horizontal plane. From a
point on the ground, the angles of elevation of the top and bottom of the flagstaff are
45° and 30° respectively. Find the height of the tower correct to one place of
decimal.
131. The horizontal distance between two trees of different heights is 60 m. The angle of
depression of the top of the first tree as seen from the top of the second tree is 45° if
the height of the second tree is 80 m, find the height of the first tree.
Graphics By :- Roshan Dhawan
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Written By :- R. K. Badhan
ACADEMY
TRIGONOMETRY
132. From the top of a building 60 m high, the angles of depression of the top and bottom
of a vertical lamp post are observed to be 30° and 60° respectively. Find
(a) The horizontal distance between the lamp post and the building.
(b) The difference between the heights of the building and the lamp post.
133. From the top of a cliff 100 m high, the angles of depression of the top and bottom of
a tower are observed to be 30° and 45° respectively. Find the height of the tower.
134. There are two temples one of each bank of a river, just opposite to each other. One
temple is 40 m high. As observed from the top of this temple. the angles of
depression of the top and the foot of the other temple are 12° 30` and 21° 98`
respectively. Find the meters, the width of the river and height of the other temple.
135. PQ is a post of given height ‘a’ and AB is a tower at some distance,  and  are the
angles of elevation of B, the top of the tower at P and Q respectively. Find the height
of the tower and its distance from the post.
136. The angles of elevation of the top of a tower from two points a and b from the base
and in the same straight line with it are complementary. Prove that the height is
ab .
137. The angle of elevation of a Jet plane from a point P on the ground is 60°. After a
flight of 15 second, the angle of elevation changes to 30°. If the Jet plane is flying at
a constant height of 1500 3 m, find the speed of the Jet plane.
138. A man on the top of a vertical tower observes a car moving at a uniform speed
coming directly towards it. If it takes 12 minutes for the angle of depression to
change from 30° to 45° how soon after this, will the car reach the tower.
139. At a point on level ground the angle of elevation of a vertical tower is found to be
such that its tangent is 5/12. On walking 192 m towards the tower, the tangent of the
angle is found to be 3/4. Find the height of the tower.
140. The length of a string between a kite and a point on the ground is 90 m. If the string
makes an angle  with the level ground and tan   15/ 8 , find the height of the
kite. Assume there is no slack in the string.
141. The length of a string between a kite and a point on the roof on a building 10 m high
is 180 m. If the string makes an angle h with the level ground such that tan   4/ 3 ,
how high is the kite from the ground? Assume there is no slack in the string.
142. A man on the deck of a ship is 16 m above water level. He observes that the angle of
elevation of the top of a cliff is 45° and the angle of depression of the base is 30°.
Calculate the distance of the cliff from the ship and the height of the cliff.
143. A tower in a city is 150 m high and a multistoried hotel at the city center is 20 m
high. The angle of elevation of the top of the tower at the top of the hotel is 5°. A
Graphics By :- Roshan Dhawan
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Written By :- R. K. Badhan
ACADEMY
144.
145.
146.
147.
148.
TRIGONOMETRY
building, h meters high, is situated on the straight road connecting the tower with
the city centre at a distance of 1.2 km from the tower. Find the value of ‘h’ if the top
of the hotel, the top of the building and top of the tower are in the straight line. Also
find the distance of the tower from the city centre.
(Use tan5  0.0875, tan85  11.43) .
The line joining the top of a hill to the foot of the hill makes angle of 30° with the
horizontal through the foot of the hill. There is one temple at the top of the hill and a
guest house half way from the foot to the top. The tops of the temple and of the
guest house both make an elevation of 45° at the foot of the hill. If the guest house is
1 km away from the foot of the hill along the hill, find the height of the guest house
and the temple.
A carpenter makes stools for elevation with a square top of side 0.5 m and at a
height 1.5 m above the ground. Also each leg is inclined at an angle of 75° to the
ground. Find the length of each leg and also the length of two steps to be put at
equal distances in meters.
A boy is standing on the ground and flying a kite with 100 m string at an elevation
of 30°. Another boy is standing on the roof of a 20 m high building and is flying his
kite at an elevation of 45°. Both the boys are on opposite sides of both the kites. Find
the length of the string such that the two kites meet.
From the top of a light house, the angle of depression of two ships on the opposite
sides of the light house are observed to be  and  . If the height of the light house
be h meters and if the line joining the ships passes through the foot of the light
 tan   tan  
house, show that the distance between the ships is h 
.
 tan  tan  
A vertical tower stands on a horizontal plane and is surmounted by a vertical
flagstaff of height h. At a point on a plane the angle of elevation of the bottom of the
flagstaff is  and that of the top of the flagstaff is  . Prove the height of the flagstaff
is h tan  /  tan tan   .
149. If the angle of elevation of a cloud from a point h meters above a lake is  and the
angle of depression of its reflection in the lake is  , prove that the height of the cloud
h  tan   tan  
is
.
 tan   tan  
150. From a window (h meters 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 the opposite
Graphics By :- Roshan Dhawan
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Written By :- R. K. Badhan
ACADEMY
TRIGONOMETRY
side of the street are  and  respectively. Show that the height of the opposite
house is h 1  tan  cot   , h = 60 m,   60,   45 .
151. Two stations due south of a leaning tower which leans towards the north are at
distances a and b from its foot. If ,  be the elevation of the top of the tower from
these stations, prove that its inclination  to the horizontal is given by
 bcot   acot  
cot   
.
ba


152. A round balloon of radius ‘a’ subtends an angle  at the eye of the observer while
the angle of elevation of its center is  . Prove that the height of the center of the

balloon is : a sin  cosec .
2
153. The angle of elevation of a cliff from a fixed point A is  . After going up distance of
k meters towards the top of the cliff at an angle of  , it is found that angle of
elevation is  . Show that the height of the cliff is k  cos   sin  cot   /  cot   cot   .
154. At the foot of a mountain the elevation of its summit is 45° after ascending 1000m
towards the mountain up a slope of 30° inclination the elevation is found to be 60°.
Find the height of the mountain.
155. If the angle of elevation of a cloud from a point h meters above a lake is  and the
angle of depression of its reflection in the lake is  . Prove that the distance of the
cloud from the point of observation is 2 hsec  /  tan   tan   .
156. From an aeroplane vertically above a straight horizontal road, the angles of
depression of two consecutive milestone on opposite sides of the aeroplane are
observed to be  and  . Show that the height of aeroplane above the road is tan 
tan  / (tan  + tan  ).
157. A ladder rests against a wall at angle  at a point A in the plane of its base and the
angle of depression of the foot of the tower at a point b meters just above A is  .
Prove that the height of tower is b tan  cot  .
158. Find the length of the chord of a circle of a circle of radius 6 cm, subtending at the
center an angle of (i) 144°
(ii) (14.6)°
159. Find the radius of the encircle of a regular polygon of 18 sides each of length 60 cm.
160. Find the length of each side of a regular polygon of 25 sides inscribed in a circle of
radius 80 cm.
Graphics By :- Roshan Dhawan
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Written By :- R. K. Badhan
ACADEMY
TRIGONOMETRY
161. From the top of a building 15 m high, the angle of elevation of the top of a tower is
found to be 30°. From the bottom of the same building, the angle of elevation of the
top of the tower is found to be 60°. Find the height of the tower and the distance
between the tower and the building.
162. If Tn  cosn   sinn  , prove that 2T6  3T4  1  0 .
T T T T
163. If Tn  sinn   cos n , then show that: 3 5  5 7 .
T1
T3
164. The angle of elevation of the top of a tower from a point A due South of the tower
is  and from B due East of the tower is  . If AB = d, show that the height of the
d
tower is
.
2
2
cot   cot 
165. The elevation of a tower at a stadium A due North of it is  and at a station B due
AB sin  sin 
West of A is  . Prove that the height of the tower is
sin 2   sin 2 
166. The angle of elevation of a certain peak when observed from each end of a
horizontal base line of length 2a is found to be  . When observed from the midpoint of the base the angle of elevation is  . Prove that the height of the peak is
a sin  sin 
.
sin2   sin2 
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Written By :- R. K. Badhan