CBSE X Maths — Heights and Distances: Notes, Practice Questions & Worked Solutions

CBSE Class X Mathematics (041/241) · Unit V: Trigonometry

Heights and Distances

This topic applies trigonometric ratios to real situations: finding how tall something is, or how far away it is, using only an angle and one known length.

1. Angle of Elevation and Depression

The angle of elevation is the angle from the horizontal up to an object above eye level. The angle of depression is the angle from the horizontal down to an object below eye level. Both are measured from the observer’s horizontal line of sight.

Worked example — finding a height The angle of elevation of the top of a tower from a point 30 m from its base is 30°. Find the height of the tower.
h/30 = tan30° → h = 30 × (1/√3) = 10√3 m

Worked example — finding a distance from an angle of depression From the top of a 10 m building, the angle of depression of a car is 30°. Find the distance of the car from the base of the building.
Since the angle of depression from the top equals the angle of elevation from the car (alternate angles), tan30° = 10/distance → distance = 10/tan30° = 10√3 m

Worked example — two angles, one problem Two poles of equal height stand on either side of a road 80 m wide. From a point between them, the angles of elevation of their tops are 60° and 30°. Find the height of the poles and the position of the point.
Let the point be x m from the first pole. h = x·tan60° and h = (80−x)·tan30°. Setting equal: x√3 = (80−x)/√3 → 3x = 80−x → x = 20.
Height = 20√3 m ≈ 34.6 m; the point is 20 m from one pole and 60 m from the other.

Worked example — complementary angles proof The angle of elevation of the top of a tower from two points at distances 4 m and 9 m from the base (in a line) are complementary. Prove the height is 6 m.
Let the angles be θ and 90°−θ. tanθ = h/4, and tan(90°−θ) = cotθ = h/9.
Multiplying: tanθ·cotθ = 1 = (h/4)(h/9) = h²/36 → h² = 36 → h = 6 m

Common mistakes to watch for

Confusing which trigonometric ratio to use — always identify which side is opposite/adjacent to the given angle first, and draw a clear diagram.
Forgetting to subtract observer height in problems that mention a person’s eye level, not ground level.
Mixing up angle of elevation and angle of depression — remember they’re equal (alternate angles) when lines of sight are parallel to the horizontal.

Key Vocabulary Recap

Angle of elevation — angle up from horizontal to a higher object

Angle of depression — angle down from horizontal to a lower object

Syllabus Reference

This note covers CBSE Class X Mathematics Unit V, Heights and Distances: simple problems on heights and distances using angles of elevation and depression of 30°, 45°, and 60°, restricted to at most two right triangles.

Practice questions

CBSE Class X Mathematics · Unit V: Heights and Distances — Practice Question Bank (no answers)

Heights and Distances — Practice

Covers: angle of elevation/depression problems using 30°, 45°, and 60°. 10 questions.

The angle of elevation of the top of a tower from a point 30 m away from its base is 30°. Find the height of the tower.

A ladder 15 m long makes an angle of 60° with the wall. Find the height it reaches on the wall.

From the top of a 10 m high building, the angle of depression of a car on the ground is 30°. Find the distance of the car from the base of the building.

Two poles of equal height stand on either side of a road 80 m wide. From a point between them, the angles of elevation of their tops are 60° and 30°. Find the height of the poles and the position of the point.

The shadow of a tower is 40 m longer when the sun’s altitude is 30° than when it is 60°. Find the height of the tower.

From a point on the ground, the angle of elevation of the top of a 20 m tall tower is 45°. Find the distance of the point from the foot of the tower.

An observer 1.5 m tall is 28.5 m away from a 30 m tall tower. Find the angle of elevation of the top of the tower from the observer’s eye.

The angles of elevation of the top of a tower from two points at distances of 4 m and 9 m from its base (on the same side, in line) are complementary. Prove that the height of the tower is 6 m.

A kite is flying at a height of 60 m. The string is tied to a point on the ground, inclined at 60° to the ground. Find the length of the string.

Q10

From the top of a 100 m high cliff, the angles of depression of the top and bottom of a tower are 30° and 60° respectively. Find the height of the tower.

More Practice

Q11

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.

Q12

A 1.6 m tall observer is 20√3 m away from a tower, and the angle of elevation from the observer’s eye to the top of the tower is 30°. Find the height of the tower.

Q13

From the top of a tower 50 m high, the angles of depression of the top and bottom of a pole are 45° and 60°. Find the height of the pole.

Q14

The angle of elevation of a cloud from a point 60 m above a lake is 30°, and the angle of depression of its reflection in the lake is 60°. Find the height of the cloud.

Q15

A man standing on the deck of a ship 10 m above sea level observes the angle of elevation of a cliff as 45° and the angle of depression of its base as 30°. Find the height of the cliff.

Q16

From a window 15 m high above the ground in a street, the angles of elevation and depression of the top and foot of another house on the opposite side are 30° and 45° respectively. Find the height of the opposite house.

Q17

Two ships are sailing on either side of a 200 m tall lighthouse. The angles of depression of the two ships from the top are 60° and 45°. Find the distance between the two ships.

Q18

An aeroplane flying at a height of 300 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.

Q19

The shadow of a tower, when the angle of elevation of the sun is 45°, is found to be 10 m longer than when the elevation was 60°. Find the height of the tower.

Q20

The angle of elevation of the top of a tower from a point on the ground is 45°. On moving 20 m towards the tower, the angle of elevation becomes 60°. Find the height of the tower.

Worked solutions

Show worked solutions

CBSE Class X Mathematics · Unit V: Heights and Distances — Solutions

Heights and Distances — Practice: Solutions

Elevation 30° from 30m away. Find height.

h = distance × tan(angle)

30 × tan30° = 30/√3

✓10√3 m ≈ 17.3 m

Ladder 15m, 60° with wall. Find height.

Height along wall = ladder × cos(angle with wall)

15 × cos60°

✓7.5 m

Depression 30° from 10m building. Find distance.

distance = height/tan(angle)

10/tan30° = 10√3

✓10√3 m ≈ 17.3 m

Poles, road 80m, angles 60° and 30°.

Let point be x from first pole: h=x tan60°=(80−x)tan30°

x√3=(80−x)/√3 → 3x=80−x → x=20

h=x tan60°

20√3

✓Height = 20√3 m ≈ 34.6 m; point is 20 m from one pole, 60 m from the other

Shadow difference 40m between 30° and 60°.

Shadow at 30°: h/tan30°=h√3. Shadow at 60°: h/tan60°=h/√3. Difference=40

h√3 − h/√3 = 40 → h(2/√3)=40

✓h = 20√3 m ≈ 34.6 m

Elevation 45°, tower 20m. Find distance.

distance = height/tan45°

20/1

✓20 m

Observer 1.5m, 28.5m from 30m tower.

Effective height above eye level = 30−1.5=28.5m

tanθ=28.5/28.5=1

✓θ = 45°

Complementary angles at 4m,9m. Prove h=6m.

tanθ=h/4, tan(90−θ)=cotθ=h/9. Multiply: tanθ×cotθ=1

1=(h/4)(h/9)=h²/36

✓h²=36, so h=6 m (proved)

Kite at 60m, string at 60° to ground.

length = height/sin(angle)

60/sin60°=60/(√3/2)

✓40√3 m ≈ 69.3 m

Q10

Cliff 100m, depressions 30°(tower top) and 60°(tower bottom).

Horizontal distance from bottom-of-tower sighting: x=100/tan60°

x=100/√3

From top-of-tower sighting: 100−AB = x tan30°, so tower height AB = 100 − x tan30°

AB=100−(100/√3)(1/√3)=100−100/3

✓Height of tower = 200/3 m ≈ 66.7 m

More Practice — Solutions

Q11

Tower 50m, elevation angles 30°,60° between tower and building feet.

Distance d=50/tan60°=50/√3

Building height = d×tan30° = (50/√3)(1/√3)

✓50/3 m ≈ 16.7 m

Q12

Observer 1.6m, 20√3m away, elevation 30°.

Height above eye level = distance×tan30°

20√3×(1/√3)=20m

✓Total height = 20+1.6 = 21.6 m

Q13

Tower 50m; depressions to pole top(45°) and bottom(60°).

Distance=50/tan60°=50/√3. Top-of-pole height above ground = 50 − distance×tan45°

50 − 50/√3

✓Pole height = 50(1−1/√3) ≈ 21.13 m

Q14

Point 60m above lake; cloud elevation30°, reflection depression60°.

Let cloud height=h, horizontal distance=d. tan30°=(h−60)/d, tan60°=(h+60)/d

(h−60)√3=(h+60)/√3 → 3(h−60)=h+60

✓2h=240 → h = 120 m

Q15

Deck 10m; cliff elevation45°, base depression30°.

Distance=10/tan30°=10√3

Height above sea = 10+10√3×tan45°=10+10√3

✓Height of cliff = 10+10√3 ≈ 27.3 m

Q16

Window 15m; elevation30°,depression45° to opposite house.

Distance=15/tan45°=15m

Height above window = 15×tan30°=15/√3=5√3

✓Total height = 15+5√3 ≈ 23.66 m

Q17

Lighthouse 200m; depressions 60°,45° to two ships.

d₁=200/tan60°, d₂=200/tan45°

d₁=200/√3, d₂=200

✓Distance between ships = d₁+d₂ ≈ 315.5 m

Q18

Plane 300m; depressions45°,60° to river banks.

d₁=300/tan45°=300, d₂=300/tan60°=100√3

width=d₁+d₂

✓Width of river ≈ 473.2 m

Q19

Shadow difference 10m between 45° and 60°.

h/tan45° − h/tan60° = 10 → h(1−1/√3)=10

h=10/(1−1/√3)=10√3/(√3−1)

✓h = 15+5√3 ≈ 23.66 m

Q20

Elevation 45° then 60° after moving 20m closer.

Let original distance=d, height=h=d (from tan45°=1). After moving 20m: h=(d−20)tan60°=(d−20)√3

d=(d−20)√3 → d(√3−1)=20√3

✓h = d = 30+10√3 ≈ 47.3 m

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