23. Chapter 23. Trigonometric Ratios of Some Special Angles NIOS Class 10 Mathematics Textbook Solutions
Chapter 23. Trigonometric Ratios of Some Special Angles NIOS Class 10 Mathematics Textbook Solutions for Exam Preparation
Chapter 23. TRIGONOMETRIC RATIOS OF SOME SPECIAL ANGLES
CHECK YOUR PROGRESS 23.1
1. Evaluate each of the following:
(i) 
(ii) 
(iii) 
(iv) 
(v) 
(vi) 
Solution: (i) We have, 
(ii) We have, 
(iii)
Solution: We have, 
(iv) We have,
(v) We have, 
(vi) We have, 
2. Verify each of the following:
(i) 
(ii) 
(iii) 
(iv) 
(v) 
Solution: (i) LHS: 
RHS
LHS = RHS verified.
(ii) LHS : 
RHS
LHS = RHS verified.
(iii) LHS 
RHS
LHS = RHS verified.
(iv) LHS: 
RHS
LHS = RHS verified.
(v) LHS : 
RHS
LHS = RHS verified.
3. If ∠A = 30° , verify each of the following:
(i) 
(ii) 
(iii) 
Solution: (i) 
LHS: 
RHS: 
LHS = RHS verified.
(ii) 
LHS: 
RHS: 
LHS = RHS verified.
(iii)
LHS: 
RHS: 
RHS = LHS Verified.
4. If A = 60° and B = 30° , verify each of the following:
(i) 
(ii) 
Solution: (i)
LHS: 
RHS: 
LHS = RHS verified.
(ii) 
LHS: 
RHS: 
LHS = RHS verified.
5. Taking 2A = 60° , find sin30° and cos30°, using 
Solution: Given, 
Now, 
And 
6. Using the formula 
, evaluate 
.
Solution: We have, 
7. If 
, 0° < A + B < 90° , A > B, find A and B.
Solution: We have, 
and 
Putting 
in (i) , we get 
8. If 
and 
, find A and B.
Solution: We have,
and 
[From (i)]
Putting 
in (i) , we get 
Therefore, 
9. In ΔPQR right angled at Q, PQ = 5 cm and ∠R = 30° , find QR and PR.
Solution: Here, PQ = 5 cm and ∠R = 30°
We have, 
And 
10. In ΔABC, ∠B = 90°, AB = 6 cm and AC = 12 cm. Find ∠A and ∠C.
Solution: Here, ∠B = 90°, AB = 6 cm and AC = 12 cm .
And 
11. In ΔABC, ∠B = 90° . If A = 30°, find the value of 
Solution: Here, 
12. If 
, find x.
Solution: We have, 
°
Choose the correct alternative for each of the following (13-15):
13. The value of 
is
(A) 2 (B) 
(C) 
(D) 
Answer: (C)
[ We have ,
]
14. If 
, then A is
(A) 30° (B) 0° (C) 60° (D) 90°
Answer: (B) 0°
[ Putting, A = 0
LHS: 
RHS: 
]
15. 
is equal to
(A) sin 60° (B) sin 30° (C) cos 60° (D) tan 60°
Answer: (A) 
[ 
]
CHECK YOUR PROGRESS 23.2
1. A ladder leaning against a vertical wall makes an angle of 60° with the ground. The foot of the ladder is at a distance of 3 m from the wall. Find the length of the ladder.
Solution: In given figure, Here, PR = The length of the ladder , PQ = 3 m and QR = height of wall
In 
, we have
Therfore, the length of the ladder is 
.
2. At a point 50 m away from the base of a tower, an observer measures the angle of elevation of the top of the tower to be 60° . Find the height of the tower.
Solution: In the given figure,
Here, AB = 50 m and BC = height of the tower
In 
, we have
Therefore, the height of the tower is 
.
3. The angle of elevation of the top of the tower is 30°, from a point 150 m away from its foot. Find the height of the tower.
Solution: In the given figure,
Here, AB = 150 m and BC = height of the tower
In , we have 
Therefore, the height of the tower is .
4. The string of a kite is 100 m long. It makes an angle of 60° with the horizontal ground. Find the height of the kite, assuming that there is no slack in the string.
Solution: In the given figure,
Here, AC = 100 m and BC = height of the tower
In , we have 
Therefore, the height of the kite is .
5. A kite is flying at a height of 100 m from the level ground. If the string makes an angle of 60° with a point on the ground, find the length of the string assuming that there is no slack in the string.
Solution: In the given figure,
Here, BC = 100 m and AC = the length of the string
In , we have
Therefore, the height of the kite is 
.
6. Find the angle of elevation of the top of a tower which is 
m high, from a point at a distance of 100 m from the foot of the tower on a horizontal plane.
Solution: In the given figure,
Here, 
and AB = 100 m
In , we have 
Therefore, the angle of elevation is 60° .
7. A tree 12 m high is broken by the wind in such a way that its tip touches the ground and makes an angle of 60° with the ground. At what height from the ground, the tree is broken by the wind?
Solution: Let the tree break at a height of h metres from the ground.
In the given figure, Here , 
In , we have
The tree is broken at a height of 5.61 m​ from the ground.
8. A tree is broken by the storm in such way that its tip touches the ground at a horizontal distance of 10 m from the tree and makes an angle of 45° with the ground. Find the height of the tree.
Solution: In given figure,
Here, AB = 10 m , BD = Height of the tree , BC = standing part of the tree , AC = CD = Broken part of the tree .
In , we have 
Again , 
and 
Therefore, the height of the tree is 24.14 m .
9. The angle of elevation of a tower at a point is 45° . After going 40 m towards the foot of the tower, the angle of elevation becomes 60° . Find the height of the tower.
Solution: In given figure,
Here, BC = the height of the tower . AD = 40 m
In 
, we have 
In , we have 
And 
Therefore, the height of the tower is 94.6 m .
10. Two men are on either side of a cliff which is 80 m high. They observe the angles of elevation of the top of the cliff to be 30° and 60° respectively. Find the distance between the two men.
Solution: In given figure,
Here, CD = 80 m , AB = the distance between the two men .
In 
, we have 
In 
, we have 
and 
Therefore, the distance between the two men is 184.97 m .
11. From the top of a building 60 m high, the angles of depression of the top and bottom of a tower are observed to be 45° and 60° respectively. Find the height of the tower and its distance from the building.
Solution: In given figure,
Here, BE = height of the building = 60 , AC = height of the tower .
i.e., AB = CD and AC = BD
In 
, we have 
In 
, we have 
And 
Therefore, the height of the tower is 25.4 m and its distance from the building is 34.6 m .
12. A ladder of length 4 m makes an angle of 30° with the level ground while leaning against a window of a room. The foot of the ladder is kept fixed on the same point of the level ground. It is made to lean against a window of another room on its opposite side, making an angle of 60° with the level ground. Find the distance between these rooms.
Solution: In given figure,
Here, AC=BC = 4 m , AB = the distance between the room.
In , we have
In , we have 
Therefore, the distance between the two men is 5.46 m .
13. At a point on the ground distant 15 m from its foot, the angle of elevation of the top of the first storey is 30°. How high the second storey will be, if the angle of elevation of the top of the second storey at the same point is 45° ?
Solution: In given figure,
Here, AB = 15 m , AB = the distance between the observer and storey , BC = the height of first storey , CD = the height of second storey .
In ∆BAC , we have
In 
, we have
The height of the second storey is 6.35 m .
14. An aeroplane flying horizontal 1 km above the ground is observed at an elevation of 60° . After 10 seconds its elevation is observed to be 30° . Find the speed of the aeroplane.
Solution: In given figure,
Here, 
, 
In 
we have , 
In 
we have ,
The speed of the aeroplane 
Therefore, the speed of the aeroplane is 115.61 m/s .
15. The angle of elevation of the top of a building from the foot of a tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60° . If the tower is 50 m high, find the height of the building.
Solution: In given figure,
Here, AB = 50 m , CD = the height of the building ,
BC = the distance between the building and the tower .
The angle elevation are 
and 
In we have
In 
we have
From (i) and (ii) , we get 
Therefore, the height of the building is 16.67 m .
TERMINAL EXERCISE
1. Find the value of each of the following:
(i) 
(ii) 
(iii) 
(iv) 
Solution: (i) W have, 
(ii) sin²45° – tan² 45° + 3(sin²90° + tan² 30°)
Solution: We have, 
(iii) We have, 
(iv) We have, 
2. Prove each of the following:
(i) 
(ii) 
(iii) 
(iv) 
Solution: (i)
LHS: 
RHS Proved.
(ii)
LHS : 
RHS Proved.
(iii)
LHS: 
RHS: 
LHS = RHS Proved.
(iv) 
LHS: 
RHS: 
LHS = RHS Proved.
3. If 
, verify that
(i) 
(ii) 
(iii) 
Solution: (i)
LHS: 
RHS: 
LHS = RHS verified.
(ii)
LHS: 
RHS: 
LHS=RHS verified.
(iii)
LHS: 
RHS: 
LHS = RHS verified.
4. If A = 60° and B = 30°, verify that
(i) sin (A + B) ≠ sin A + sin B
(ii) sin (A + B) = sin A cos B + cos A sin B
(iii) cos (A – B) = cos A cos B + sin A sin B
(iv) cos (A + B) = cos A cos B – sin A sin B
(v) 
Solution: (i) 
LHS : 
RHS : 
verified
(ii) sin (A + B) = sin A cos B + cos A sin B
LHS :
RHS:
LHS = RHS verified.
(iii) cos (A – B) = cos A cos B + sin A sin B
LHS: 
RHS: 
LHS = RHS verified.
(iv) cos (A + B) = cos A cos B – sin A sin B
LHS: 
RHS: 
LHS =RHS verified.
(v)
LHS: 
RHS: 
LHS = RHS verified.
5. Using the formula cos (A – B) = cos A cos B + sin A sin B, find the value of cos 15°.
Solution: We have, 
6. If 
and 
, 
, A > B, find A and B.
Solution: We have, 
…………… (i)
and
…………….. (ii)
Putting 
in (i) , we get 
Therefore, 
and 
.
7. An observer standing 40 m from a building observes that the angle of elevation of the top and bottom of a flagstaff, which is surmounted on the building are 60° and 45° respectively. Find the height of the tower and the length of the flagstaff.
Solution: In given figure,
Here, AB = 40 m , BC = the height of building , CD = the height of the flagstaff .
In 
, we have 
In , we have 
and 
Therefore, the height of the tower is 40 m and the length of the flagstaff is 29.2 m .
8. From the top of a hill, the angles of depression of the consecutive kilometre stones due east are found to be 60° and 30° . Find the height of the hill.
Solution: In given figure,
Here, CD = the height of the hill , AB = 1 km = 1000 m
In 
we have ,
In 
we have , 
Therefore, the height of the hill is 865 m .
9. From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45° . Find the height of the tower.
Solution: In given figure,
Here, AB = CE = height of the building = 7 m ;
BC = AE = the distance between the tower and the building ,
and 
In 
we have ,
In 
we have ,
From 
and 
we get , 
m
Therefore, the height of the tower is 19.11 m .
10. A man on the top of a tower on the sea shore finds that a boat coming towards him takes 10 minutes for the angle of depression to change from 30° to 60° . How soon will the boat reach the sea shore?
Solution: In given figure,
Here, CD = the height of the tower , time for AB = 10 minutes .
In we have ,
In we have ,
⇒BC=12AB 
The time for the distance 
Therefore, the boat will reach the sea shore in 5 minutes.
11. Two boats approach a light-house from opposite directions. The angle of elevation of the top of the lighthouse from the boats are 30° and 45° . If the distance between these boats be 100 m, find the height of the lighthouse.
Solution: In given figure,
Here , AB = 100 m , CD = the height of the lighthouse .
and 
In 
we have , 
In 
we have ,
Therefore, the height of the lighthouse is 36.63 m .
12. The shadow of a tower standing on a level ground is found to be 
m longer when the sun’s altitude is 30° than when it was 60°. Find the height of the tower.
Solution: In given figure,
Here, 
, CD = the height of the tower .
and 
In we have ,
In we have ,
Therefore, the height of the tower is 67.5 m .
13. The horizontal distance between two towers is 80 m. The angle of depression of the top of the first tower when seen from the top of the second tower is 30°. If the height of the second tower is 160 m, find the height of the first tower.
Solution: In given figure,
Here, BC = the distance between two tower, CE = 160 m , AB = height of the first tower .
So, AB = CD , AD = BC and 
In 
we have ,
Therefore, the height of the first tower is 113.76 m .
14. From a window, 10 m 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 the street are 60° and 45° respectively. Find the height of the opposite house (Take 
)
Solution: In given figure,
Here , AB = 10 m, CE = the height of the opposite house .
So, AB = CD = 10 m and AD = BC
and
In we have , 
In 
we have , 
Therefore, the height of the opposite house is 27.3 m .
15. A statue 1.6 m tall stands on the top of a pedestal from a point on the gound, the angle of elevation of the top of the statue is 60° and from the same point, the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.
Solution: In given figure :
Here, CD = 1.6 m , BC = the height of the pedestal .
AB = the distance between the ground and the foot point of the pedestal .
So, 
and 
.
In we have
In 
we have,
Therefore, the height of the pedestal is 2.18 m .
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