Heights and Distances

Q: The angle of elevation of the top of a 25 m tall tree is 45°. How far is the observer from the foot of the tree?

tan 45° = 1, so the distance equals the height: 25 m.

Q: From the top of a cliff 50 m high, the angle of depression of a buoy is 60°. Find the distance of the buoy from the foot of the cliff. (Take √(3) = 1.732.)

Distance = (50)/(tan 60°) = (50)/(√(3)) = (50√(3))/(3) = 28.87 m.

Q: A ladder 10 m long reaches a window when it makes an angle of 60° with the ground. How high is the window? (Take √(3) = 1.732.)

Height = 10 sin 60° = 5√(3) = 8.66 m.

ICSE Class 10 · Mathematics · Chapter 22

What this chapter covers

Measuring what you cannot reach
Line of sight and the horizontal
Angle of elevation
Angle of depression
Choosing the right ratio
Using trigonometric tables
Problems with one right-angled triangle
Two triangles on the same side of an object
Two triangles on opposite sides of an object
Elevation and depression from a height
Moving objects and speed
Common mistakes and how to avoid them

Key points

The line of sight joins the observer's eye to the object; the horizontal line through the eye is parallel to level ground.
The angle of elevation is measured upwards from the horizontal to the line of sight; the angle of depression is measured downwards from the horizontal.
The angle of depression from A to B equals the angle of elevation from B to A, because they are alternate angles between parallel horizontals.
Use tan or cot when the height and ground distance are involved; use sin or cos when a slant length is involved.
At 45^° the height equals the horizontal distance, since tan 45^° = 1.
In four-figure tables, add the mean difference for sine and tangent and subtract it for cosine.
To find an angle from a ratio, locate the nearest smaller table entry, then read the extra minutes from the mean difference columns.
Add the observer's eye height to the height found from the triangle when it is given.
Points on the same side of an object are separated by the difference of their distances from its foot; points on opposite sides, by the sum.
Speed equals distance moved (from two triangles) divided by time; multiply m/s by (18)/(5) for km/h.
The syllabus limits problems to 2-D situations with at most two right-angled triangles.

Questions answered

Define the angle of elevation and the angle of depression.

The angle of elevation of an object is the angle between the horizontal line through the observer's eye and the line of sight when the object is above the horizontal; the angle of depression is the same angle when the object is below the horizontal.

Explain, with a reason, why the angle of depression of a boat seen from a cliff top equals the angle of elevation of the cliff top seen from the boat.

The horizontal through the cliff top is parallel to the sea surface, and the line of sight is a transversal. The two angles are alternate angles between parallel lines, so they are equal.

A man walks towards a tower. Does the angle of elevation of its top increase or decrease? Give a reason.

It increases. tanθ = (h)/(d); as d decreases, (h)/(d) increases, and tanθ increases with θ between 0^° and 90^°.

The angle of elevation of the top of a 25 m tall tree is 45^°. How far is the observer from the foot of the tree?

tan 45^° = 1, so the distance equals the height: 25 m.

From the top of a cliff 50 m high, the angle of depression of a buoy is 60^°. Find the distance of the buoy from the foot of the cliff. (Take √(3) = 1.732.)

Distance = (50)/(tan 60^°) = (50)/(√(3)) = (50√(3))/(3) = 28.87 m.

A ladder 10 m long reaches a window when it makes an angle of 60^° with the ground. How high is the window? (Take √(3) = 1.732.)

Height = 10 sin 60^° = 5√(3) = 8.66 m.

Using a table in which tan 26^° 30' = 0.4986 and the mean difference for 3' in that row is 11, find tan 26^° 33'.

Tangent increases with the angle, so add: 0.4986 + 0.0011 = 0.4997.

Using a table in which cos 40^° 12' = 0.7638 and the mean difference for 2' in that row is 4, find cos 40^° 14'.

Cosine decreases as the angle increases, so subtract: 0.7638 - 0.0004 = 0.7634.

Past board exam questions

ICSE 2027 sample paper · 5 marks

Draw the necessary diagram for this question. A man on the top of a lighthouse observes the angle of depression of two ships on the opposite sides of the lighthouse as 32° and 58° respectively. If the height of the lighthouse is 120 m, find the distance between the two ships. Give your answer correct to the nearest meter.

Show answer

Let AD = 120 m be the lighthouse (A at the top, D at the foot), with ships B and C on opposite sides. Angles of depression 32° (at B) and 58° (at C), so ∠ABD = 32° and ∠ACD = 58°; ∠BAD = 58° and ∠CAD = 32°. In △ABD: (BD)/(AD) = tan 58° ⇒ BD = 120 × 1.6003 = 192.04 m In △ACD: (CD)/(AD) = tan 32° ⇒ CD = 120 × 0.6249 = 74.99 m BC = BD + CD = 120 × 2.2252 ≈ 267.0 m = 267 m.

ICSE 2026 sample paper · 5 marks

Draw the necessary diagram for this question. A man on the top of a lighthouse observes the angle of depression of two ships on the opposite sides of the lighthouse as 30° and 50° respectively. If the height of the lighthouse is 80m, find the distance between the two ships. Give your answer correct to the nearest meter. (Use Mathematical Tables for this Question)

Show answer

Diagram: lighthouse AB = 80 m (A at the top, B at the foot); ship C on one side with angle of depression 30°, ship D on the other side with angle of depression 50°; C, B, D on one horizontal line. So ∠ACB = 30° and ∠ADB = 50°. In ∆ABC: (AB)/(BC) = tan 30^° ⇒ BC = 80√(3) = 80 × 1.7321 = 138.57 m. In ∆ABD: (AB)/(BD) = tan 50^° ⇒ BD = 80 cot 50^° = 80 tan 40^° = 80 × 0.8391 = 67.13 m. Distance between the ships CD = BC + BD = 80(1.7321 + 0.8391) = 80 × 2.5712 = 205.70 m ≈ 206 m.

ICSE 2026 board paper · 5 marks

A flagpole is erected at the top of a building. The angle of elevation of the top and foot of the flagpole from a point 100 m away, on the same level as that of the foot of the building, are 33° and 31° respectively. Find the height of the flagpole. Give your answer correct to the nearest metre. (Use Mathematical Tables for this question.)

Show answer

Let A be the point, B the foot of the building, C the top of the building (foot of the flagpole) and D the top of the flagpole; AB = 100 m. In ΔABC: tan 31^° = (BC)/(100) ⇒ BC = 100 × 0.6009 = 60.09 m. In ΔABD: tan 33^° = (BD)/(100) ⇒ BD = 100 × 0.6494 = 64.94 m. Height of the flagpole CD = BD − BC = 64.94 − 60.09 = 4.85 m ≈ 5 m.

ICSE 2025 sample paper · 1 mark

In triangle PQR, ∠PQR = 90°, QR = 9 cm and ∠QPR = 60°. The length of PR is:

3√(3) cm
6√(3) cm
9√(3) cm
18 cm

Show answer

6√(3) cm

ICSE 2025 compartment paper · 1 mark

Assertion (A): If the length of shadow of a person is equal to his height, then the angle of elevation of the sun is 45°. Reason (R): For any right-angled triangle, tanθ = (Perpendicular)/(Base)

(A) is true, (R) is false.
(A) is false, (R) is true.
Both (A) and (R) are true, and (R) is the correct reason for (A).
Both (A) and (R) are true, and (R) is the incorrect reason for (A).

Show answer

Both (A) and (R) are true, and (R) is the correct reason for (A).

ICSE 2025 sample paper · 5 marks

A vertical tower standing on a horizontal plane is surmounted by a vertical flagstaff. At a point 100m away from the foot of the tower, the angle of elevation of the top and bottom of the flagstaff are 54° and 42° respectively. Find the height of the flagstaff. Give your answer correct to nearest metre.

Show answer

Let A be the foot of the tower, B the top of the tower (bottom of the flagstaff), F the top of the flagstaff and P the point on the ground with PA = 100 m. In ΔPAB: (AB)/(PA) = tan 42° ⇒ AB = 100 × 0.9004 = 90.04 m. In ΔPAF: (AF)/(PA) = tan 54° ⇒ AF = 100 × 1.3764 = 137.64 m. Height of flagstaff BF = 137.64 − 90.04 = 47.60 m ≈ 48 m.

ICSE 2025 compartment paper · 5 marks

The angle of elevation of the top of a hill from the foot of a tower at B is 50°. The angle of elevation of the top of the tower 100 m high from the foot of the hill at C is 35°. (The tower AB, with A at its top, stands at B and the hill CD, with D at its top, stands at C; both are vertical on the level ground BC. ∠DBC = 50° and ∠ACB = 35°.) Find the: (a) horizontal distance BC between the Hill and the Tower. (b) height CD of the Hill. (Take tan50° = 1.20) (c) time taken by a cyclist to cover the distance BC, cycling at 20 m/sec.

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Tower AB = 100 m stands at B; hill CD stands at C; B and C are on level ground. (a) In right ΔABC: tan 35^° = (AB)/(BC) ⇒ BC = (100)/(tan 35^°) = (100)/(0.7002) ≈ 142.82 m. (b) In right ΔDCB: tan 50^° = (CD)/(BC) ⇒ CD = 142.82 × 1.20 ≈ 171.38 m. (c) Time = (142.82)/(20) ≈ 7.14 s.

ICSE 2024 sample paper · 1 mark

An observer at point E, which is at a certain distance from the lamp post AB, finds the angle of elevation of top of lamp post from positions C, D and E as α, β and γ. It is given that B, C, D and E are along a straight line, in that order (C is nearest to the foot B and E is farthest). Which of the following condition is satisfied?

tanα > tan β
tan β < tan γ
tan γ > tan α
tan α < tan β

Show answer

tanα > tan β