Heights and Distances Class 10 Concise Exe-22A ICSE Maths Selina Solutions Ch-22. In this article you would learn how to solve problems / questions on Angles of Depression and Elevation. Visit official website CISCE for detail information about ICSE Board Class-10 Mathematics.

Heights and Distances Class 10 Concise Exe-22A ICSE Maths Selina Solutions Ch-22

Board	ICSE
Publications	Selina
Subject	Maths
Class	10th
Chapter-22	Heights and Distances
Writer	R.K. Bansal
Exe-22A	Angles of Depression and Elevation.
Edition	2025-2026

Solved Questions / Problems on Angles of Depression and Elevation

Class 10 Concise Exe-22A ICSE Maths Selina Solutions Ch-22 Heights and Distances

Que-1: The height of a tree is √3 times the length of its shadow. Find the angle of elevation of the sun.

Sol:  Let the length of the shadow of the tree be x m
∴ Height of the tree = √3 × 𝑚
If θ is the angle of elevation of the sun, then
tan⁡𝜃 = √3⁢𝑥/𝑥
= √3
= tan 60°
∴ θ = 60°

Que-2: The angle of elevation of the top of a tower from a point on the ground and at a distance of 160 m from its foot, is found to be 60^o. Find the height of the tower.

Sol: Let AB be the tower and C is the point which is 160 m away from the foot of the tower, i.e CB = 160 m
Let height of the tower be h
Now in right ΔABC, we have
tan⁡𝜃 = 𝐴⁢𝐵/𝐵⁢𝐶
⇒ tan⁡60° = ⁢ℎ/160
∴ h = 160 × tan 60°
= 160 × √3 𝑚
= 160 × (1.732) m
= 277.12 m
∴ Required height of the tower = 277.12 m

Que-3: A ladder is placed along a wall such that its upper end is resting against a vertical wall. The foot of the ladder is 2.4 m from the wall and the ladder is making an angle of 68^o with the ground. Find the height, upto which the ladder reaches.

Sol: Let the height upto which the ladder reaches be h m.
Given that angle of elevation is 68°
tan⁡ 68° = ⁢ℎ/2.4
⇒ 2.475 = ℎ/2.4
∴ h = 2.475 × 2.4 = 5.94 m
So, the ladder reaches upto a height of 5.94 m

Que-4: Two persons are standing on the opposite sides of a tower. They observe the angles of elevation of the top of the tower to be 30^o and 38^o respectively. Find the distance between them, if the height of the tower is 50 m.

Sol: Two persons A and B are standing on the opposite side of the tower TR and height of tower TR = 50 m and angles of elevation with A and B are 30° and 38° respectively.
Let AR = x and RB = y
Now in right ΔTAR, we have
tan⁡𝜃 = 𝑇⁢𝑅/𝐴⁢𝑅
⇒ tan⁡ 30° = ⁢50/𝑥
⇒ 1/√3 = 50/𝑥
∴ 𝑥 = 50⁢√3 = 86.60 𝑚
Again in right ΔTRB, we have
tan⁡ 38° = ⁢50/𝑦
⇒ y tan 38° = 50
𝑦 = 50/tan⁡38°
= 50/0.7813
= 63.99
or 64.00 m   …(i)
∴ Distance between A and B
= x + y
= 86.60 + 64.00
= 150.6 m

Que-5: A kite is attached to a string. Find the length of the string, when the height of the kite is 60 m and the string makes an angle 30^o with the ground.

Sol:  Let the length of the rope be x m
sin⁡30° = ⁢60/𝑥
⇒ 1/2 = 60/𝑥
∴ x = 120 m
So the length of the rope is 120 m

Que-6: A boy, 1.6 m tall, is 20 m away from a tower and observes the angle of elevation of the top of the tower to be (i) 45^o, (ii) 60^o. Find the height of the tower in each case.

Sol: Let the height of the tower be h m.
(i) Here, θ = 45°
∴ tan⁡45° = (⁢ℎ−1.6)/20
⇒ 1 = (ℎ−1.6)/20
∴ h = 21.6 m
So, height of the tower is 21.6 m

(ii) Here θ = 60°
∴ tan⁡60° = (⁢ℎ−1.6)/20
⇒ √3 = (ℎ−1.6)/20
∴ h = 20 × 1.732 + 1.6 = 36.24 m
So, height of the tower is 36.24 m.

Que-7: The upper part of a tree, broken over by the wind, makes an angle of 45^o with the ground and the distance from the root to the point where the top of the tree touches the ground is 15 m. What was the height of the tree before it was broken?

Sol:  In ΔABC,
∴ tan⁡45° = ⁢Perpendicular/Base
1 = 𝐴⁢𝐵/𝐵⁢𝐶
AB = BC = 15 meters
In right angle triangle ABC,
AC² = AB² + BC²
AC² = 15² + 15²
AC² = 225 + 225
AC² = 450
AC = √450
= 15⁢√2
Height of tree = AB + AC
= 15 +15⁢√2
= 15 + 21.21
= 36.21 m
Hence, the height of the tree before it was broken = 36.21 metres.

Que-8: The angle of elevation of the top of an unfinished tower at a point distance 80 m from its base is 30^o. How much higher must the tower be raised so that its angle of elevation at the same point may be 60^o?

Sol:  Let AB be the unfinished tower and C be the top of the tower when finished, Let P be a point 80 m from the foot A.
In ΔBAP,
tan⁡30° = ⁢𝐴⁢𝐵/𝐴⁢𝑃
⇒1/√3 = 𝐴⁢𝐵/80
⇒ 𝐴⁢𝐵  = 80/√3 = 46.19 𝑚
In ΔCAP,
tan⁡60° = ⁢𝐴⁢𝐶/𝐴⁢𝑃
⇒ √3 = 𝐴⁢𝐶/80
⇒𝐴⁢𝐶 = 80⁢√3 = 138.56 𝑚
Therefore, the tower must be raised by (138.56 – 46.19) m = 92.37 m

Que-9: At a particular time, when the sun’s altitude is 30^o, the length of the shadow of a vertical tower is 45 m. Calculate
(i) the length of the tower.
(ii) the length of the shadow of the same tower, when the sun’s altitude is
(a) 45^o (b) 60^o

Sol: Shadow of the tower = 45 m and angle of elevation = 30°
(i) Let AB be the tower and BC is its shadow.
∴ CB = 45 m.
Now in right ΔABC,
tan𝜃 = 𝐴⁢𝐵/𝐵⁢𝐶
⇒tan⁡30° = ⁢𝐴⁢𝐵/45
⇒1/√3 = 𝐴⁢𝐵/45
⇒ 𝐴⁢𝐵 = 45/√3⁢𝑚.
∴ 𝐴⁢𝐵 = (45×√3) / (√3×√3)
= 45⁢√3/3
= 15⁢√3 𝑚.
= 15 (1.732)
= 25.980
= 25.98 m

(ii) In second case,
(a) Angle of elevation = 45°
And height of tower = 25.98 m
or 15⁢√3 𝑚
tan⁡45° = ⁢𝐴⁢𝐵/𝐷⁢𝐵
⇒𝐴⁢𝐵/𝐷⁢𝐵 = 1
∴ DB = AB = 25.98 m.

(b) Angle of elevation = 60°
And height of tower = 25.98 m
or 15⁢√3 𝑚.
Let shadow of the tower DB = x m
∴ tan⁡60° = ⁢𝐴⁢𝐵/𝐷⁢𝐵
⇒ √3 = 15⁢√3 / 𝐷⁢𝐵
⇒ 𝐷⁢𝐵 = 15⁢√3/√3 = 15 𝑚.
Hence length of shadow = 15 m.

Que-10: Two vertical poles are on either side of a road. A 30 m long ladder is placed between the two poles. When the ladder rests against one pole, it makes angle 32^o24′ with the pole and when it is turned to rest against another pole, it makes angle 32^o24′ with the road. Calculate the width of the road.

Sol:   

Let AB the ladder and  ∠ABP = 32°24 .
PQ = PB + BQ = ?
In ΔABP = sin 32°24 = 𝑃⁢𝐵/𝐴⁢𝐵
PB = 30 × 0.56
PB = 16.08 m
In ΔCBQ
cos 32°24 = 𝐵⁢𝑄/𝐵⁢𝐶
BQ = 30 × 0.844
= 25.32
PQ = 16.08 + 25.32
⇒ PQ = 41.4 m

Que-11: Two climbers are at points A and B on a vertical cliff face. To an observer C, 40m from the foot of the cliff, on the level ground, A is at an elevation of 48^o and B of 57^o. What is the distance between the climbers?

Sol: 

Let P be the foot of the cliff on level ground.
Then, ∠ACP = 48° and ∠BCP = 57°
∴  𝐵⁢𝑃/𝑃⁢𝐶 = tan⁡57°
⇒ BP = 40 × 1.539 = 61.57 m
Also, 𝐴⁢𝑃/𝑃⁢𝐶 = tan⁡48°
⇒ AP = 40 × 1.110 = 44.4 m
Hence, distance between the climbers = AB = BP – AP = 17.17 m

Que-12: A man stands 9 m away from a flag-pole. He observes that angle of elevation of the top of the pole is 28^o and the angle of depression of the bottom of the pole is 13^o. Calculate the height of the pole.

Sol: 

Let AB be the man and PQ be the flag pole
Given, AR = 9 m
Also, ∠PAR = 28° and ∠QAR = 13°
∴ 𝑃⁢𝑅/𝐴⁢𝑅 = tan⁡28°
⇒ PR = 9 × 0.532 = 4.788 m
Also, 𝑅⁢𝑄/𝐴⁢𝑅 = tan⁡13°
⇒ RQ = 9 × 0.231 = 2.079 m
Hence, height of the pole = PR + RQ = 6.867 m

Que-13: From the top of a cliff 92 m high, the angle of depression of a buoy is 20^o. Calculate, to the nearest metre, the distance of the buoy from the foot of the cliff.

Sol: 

Let AB be the cliff and C be the buoy.
Given, AB = 92 m
Also, ∠ACB = 20°
∴𝐴⁢𝐵/𝐵⁢𝐶 = tan⁡20°
⇒𝐵⁢𝐶 = 92/0.3640 = 252.7 𝑚 ≈ 253 𝑚
Hence, the buoy is at a distance of 253 m from the foot of the cliff.

–: Heights and Distances Class 10 Concise Exe-22A ICSE Maths Selina Solutions Ch-22 :–

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