MathsICSE Class 10 · Heights and Distances

Angle of Depression: Diagrams, Alternate Angles and Solved Problems (ICSE Class 10)

Q: What is the angle of depression?

It is the angle between the horizontal line through the observer's eye and the line of sight to an object below the observer, measured downwards from the horizontal.

Q: Is the angle of depression equal to the angle of elevation?

Yes. The angle of depression of B seen from A equals the angle of elevation of A seen from B, because they are alternate angles between the parallel horizontal lines.

Q: How do I find the distance of a boat from a cliff using the angle of depression?

Divide the height of the cliff by the tangent of the angle of depression: d = h ÷ tan θ. Use the four-figure table value and round only the final answer.

Updated 15 October 2026 · Fact-checked

The angle of depression is the angle between the horizontal line through the observer's eye and the line of sight down to an object below. Because that horizontal is parallel to the ground, it equals the angle of elevation at the object, so the ground distance is $d = \frac{h}{\tan\theta}$.

Understand Angle of Depression

The CISCE 2027 syllabus names angles of elevation and depression together, in 2-D problems solved with trigonometric tables. Diagram errors cost marks in depression questions (2025 Q6(ii)).

Definition. An observer at a height looks down at an object. The angle of depression is measured from the horizontal line through the observer's eye, downwards to the line of sight. It is never measured from the vertical wall, tower or cliff.

The key fact: alternate angles. The horizontal through the observer is parallel to the level ground. The line of sight is a transversal, so the angle of depression at the top equals the angle of elevation at the object.

This moves the angle into the right triangle on the ground, where the right angle is at the foot of the building. Then:

height $h$ and depression $θ$ give the ground distance $d = \frac{h}{t a n θ} = h cot θ$ ;
distance $d$ and depression $θ$ give the height $h = d tan θ$ .

Two objects seen from one height.

Same side, in line with the foot: the nearer object has the larger depression. The gap is $h cot θ_{small} - h cot θ_{large}$ .
Opposite sides: add the two distances.

Moving objects. A boat sailing away from a cliff makes the depression decrease; a car approaching a tower makes it increase. Distance moved $=$ difference of the two ground distances; speed $=$ distance $\div$ time ( $\times 3.6$ for km/h).

Eye height. If the observer stands on a platform, the height in the triangle is from the ground to the eye: platform height plus eye height, when both are given.

Depression inside a two-object figure. When an observer on one building looks down at the top of a shorter tower (as in 2025 Q6(ii)), the right triangle is formed with a horizontal line drawn from the tower top across to the building. That case is on the page on two vertical objects.

Angle of Depression: formulas to remember

Alternate angles: $depression at A of B = elevation at B of A$; Horizontal at the eye is parallel to the ground.
Distance from height: $d = \frac{h}{tan θ} = h cot θ$; Most common depression step.
Two objects, same side: $d = h cot θ_{1} - h cot θ_{2}, θ_{1} < θ_{2}$; Larger depression for the nearer object.
Two objects, opposite sides: $d = h cot θ_{1} + h cot θ_{2}$; The observer is between them.
Speed: $v = \frac{d}{t} m/s = \frac{3.6 d}{t} km/h$; For a boat or car whose depression changes.

How to solve Angle of Depression questions

Draw the horizontal at the eye first, mark the depression below it, then carry the angle down to the ground by alternate angles.

1Draw the building, cliff or tower vertically, with the ground horizontal and a right angle at the foot.
2Draw a dashed horizontal line through the observer's eye, parallel to the ground.
3Mark the angle of depression between this horizontal and the line of sight, below the horizontal.
4Write the same angle at the object on the ground (alternate angles) and state the reason.
5In the ground triangle, use $tan θ = \frac{height}{distance}$ ; for two objects, add or subtract the distances.
6Substitute table values, keep four figures and round only the final answer.

Angle of Depression in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Spotting a car from a balcony

Suppose you are on a balcony 14 m above a straight road and see your friend's car parked at an angle of depression of $2 0^{\circ}$ .

How far along the ground is the car from the foot of your building, to 2 decimal places? ( $tan 2 0^{\circ} = 0.3640$ )

Try it first, then show the solutionSolution

The depression at the balcony equals the elevation at the car: $2 0^{\circ}$ .
$tan 2 0^{\circ} = \frac{14}{d}$ , so $d = \frac{14}{0.3640} \approx 38.46$ m.

Answer: About 38.46 m

Memory hook: Look down, then carry the angle down to the ground.

A pilot lining up with a runway

Imagine a pilot flying level at 2000 m who sees the start of the runway at an angle of depression of $1 2^{\circ}$ .

What is the horizontal distance to the runway, in km to 2 decimal places? ( $tan 1 2^{\circ} = 0.2126$ )

Try it first, then show the solutionSolution

$d = \frac{2000}{t a n 1 2 ^{\circ}} = \frac{2000}{0.2126} \approx 9407.3$ m.
In kilometres, $d \approx 9.41$ km.

Answer: About 9.41 km

Memory hook: Small depression, long way off.

Tricks to remember Angle of Depression

Short memory aids. Say each one aloud once, then use it on a question the same day.

Depression drops by Z
The horizontal at the top, the ground and the line of sight make a Z shape: the two angles inside the Z are equal.
Use it when: Moving the angle from the observer to the object.
Horizontal first, always
No horizontal line at the eye, no angle of depression. Draw the dashed line before marking anything.
Use it when: Avoiding the angle measured from the wall.
Near means steep
The closer the object to the foot, the steeper you look down: larger depression.
Use it when: Placing two boats or cars correctly.

Angle of Depression: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Depression with a standard angle · 1 mark · MCQBoard level

From the top of a building 30 m high, the angle of depression of a point $P$ on the level ground is $3 0^{\circ}$ . The distance of $P$ from the foot of the building is:
(a) $103$ m
(b) 30 m
(c) $303$ m
(d) 60 m

Try it first, then show the solutionSolution

Elevation at $P$ is $3 0^{\circ}$ (alternate angles).
$d = \frac{30}{t a n 3 0 ^{\circ}} = 303$ m.

Answer: (c)

Example 2: One object, table value · 3 marks · Short answerBoard level

From a window 16 m above the ground, the angle of depression of a parked scooter is $3 3^{\circ}$ . Find the distance of the scooter from the foot of the building, to 2 decimal places. ( $tan 3 3^{\circ} = 0.6494$ )

Try it first, then show the solutionSolution

Draw the horizontal at the window; the depression $3 3^{\circ}$ equals the elevation of the window from the scooter (alternate angles).
In the right triangle at the foot, $tan 3 3^{\circ} = \frac{16}{d}$ .
$d = \frac{16}{0.6494} \approx 24.64$ m.

Answer: ≈ 24.64 m

Example 3: Two boats on opposite sides · 4 marks · Short answerHarder board level

From the top of a hill 120 m high, the angles of depression of two boats on opposite sides of the hill, in line with its foot, are $3 0^{\circ}$ and $4 5^{\circ}$ . Find the distance between the boats, to 2 decimal places.

Try it first, then show the solutionSolution

Alternate angles give elevations $3 0^{\circ}$ and $4 5^{\circ}$ at the boats.
First boat: $d_{1} = \frac{120}{t a n 3 0 ^{\circ}} = 1203$ m. Second boat: $d_{2} = \frac{120}{t a n 4 5 ^{\circ}} = 120$ m.
Opposite sides, so the distance $= 1203 + 120 = 120 (3 + 1) \approx 327.85$ m.

Answer: ≈ 327.85 m

Example 4: Two villages on the same side · 4 marks · Short answerHarder board level

An aircraft flying level at 1500 m sees two villages straight ahead on the same side, in line with the point directly below it, at angles of depression of $3 9^{\circ}$ and $2 4^{\circ}$ . Find the distance between the villages, to the nearest metre. ( $tan 3 9^{\circ} = 0.8098$ , $tan 2 4^{\circ} = 0.4452$ )

Try it first, then show the solutionSolution

The larger depression ( $3 9^{\circ}$ ) belongs to the nearer village.
Nearer: $d_{1} = \frac{1500}{0.8098} \approx 1852.31$ m. Farther: $d_{2} = \frac{1500}{0.4452} \approx 3369.27$ m.
Same side, so subtract: $3369.27 - 1852.31 \approx 1516.96$ m.
To the nearest metre, 1517 m.

Answer: 1517 m

Example 5: Boat sailing away from a cliff · 5 marks · Long answerHOTS / competency

From the top of a cliff 80 m above sea level, a man watches a boat sailing directly away from the foot of the cliff. The angle of depression changes from $5 2^{\circ}$ to $3 6^{\circ}$ in 12 seconds. Find the distance travelled, to 2 decimal places, and the speed of the boat in km/h, to 1 decimal place. ( $tan 5 2^{\circ} = 1.2799$ , $tan 3 6^{\circ} = 0.7265$ )

Try it first, then show the solutionSolution

Moving away, so the depression decreases: first at $5 2^{\circ}$ , then at $3 6^{\circ}$ .
First distance $= \frac{80}{1.2799} \approx 62.505$ m; second distance $= \frac{80}{0.7265} \approx 110.117$ m.
Distance travelled $= 110.117 - 62.505 \approx 47.61$ m.
Speed $= \frac{47.612}{12} \approx 3.968$ m/s $= 3.968 \times 3.6 \approx 14.3$ km/h.

Answer: ≈ 47.61 m; ≈ 14.3 km/h

Example 6: Case study: a watchtower in a reserve · 5 marks · Case-basedHOTS / competency

A forest guard stands on a watchtower platform 24 m above level ground; her eyes are 1.5 m above the platform. She sees an elephant walking straight towards the foot of the tower at an angle of depression of $1 8^{\circ}$ ; later the angle is $3 5^{\circ}$ . ( $tan 1 8^{\circ} = 0.3249$ , $tan 3 5^{\circ} = 0.7002$ )
(a) Find the two distances of the elephant from the foot of the tower, to 2 decimal places.
(b) How far did the elephant walk?
(c) At a steady 2 m/s, how much longer will it take to reach the foot of the tower, to 1 decimal place?
(d) A student used 24 m as the height. What first distance would that give, and why is it wrong?

Try it first, then show the solutionSolution

(a) Eye height $= 24 + 1.5 = 25.5$ m. $d_{1} = \frac{25.5}{0.3249} \approx 78.49$ m; $d_{2} = \frac{25.5}{0.7002} \approx 36.42$ m.
(b) $78.486 - 36.418 \approx 42.07$ m.
(c) Time $= \frac{36.418}{2} \approx 18.2$ s.
(d) $\frac{24}{0.3249} \approx 73.87$ m. The angle is measured at her eye, so the vertical side of the triangle is the eye height, 25.5 m, not the platform height.

Answer: (a) ≈ 78.49 m, ≈ 36.42 m (b) ≈ 42.07 m (c) ≈ 18.2 s (d) ≈ 73.87 m; wrong height used

Where students lose marks in Angle of Depression

The slips examiners see most often, and what to write instead.

Mistake: Depression measured from the vertical
The angle of depression starts at the horizontal through the eye; measuring from the tower gives the complement.
Wrong working
Marking $3 5^{\circ}$ between the tower and the line of sight
Right working
Marking $3 5^{\circ}$ between the dashed horizontal and the line of sight
Do this: Draw the dashed horizontal first, then the angle below it.
Usually costs: The CISCE examiners' suggestions for 2025 Q6(ii) ask for labelled diagrams with the depression marked from the horizontal.
Mistake: Inverted ratio
In the ground triangle the height is opposite the angle and the distance is adjacent.
Wrong working
$tan 1 4^{\circ} = \frac{100}{E P}$
Right working
$tan 1 4^{\circ} = \frac{E P}{100}$
Do this: Write 'opposite' and 'adjacent' on the triangle before the ratio.
Usually costs: Reported in the CISCE examiners' comments on 2025 Q6(ii); the heights then come out wrong.
Mistake: Misreading the tables or rounding in the middle
Table values have four figures; rounding early or reading the wrong column changes the answer.
Wrong working
Rounding the first distance before subtracting the second
Right working
Keep each distance to three or four decimals; round the difference at the end
Do this: If the question gives a rounding rule for table values (2025 Q6(ii) asked for two decimals), follow it exactly; otherwise keep four figures.
Usually costs: Reported in the CISCE examiners' comments on 2025 Q6(ii): the accuracy marks.
Mistake: Ignoring eye height
The angle is measured at the eye, not at the floor.
Wrong working
Using the platform height alone when the eye height is given
Right working
Height in the triangle = platform + eye height
Do this: Add every height given above the ground up to the eye.
Usually costs: The final answer.

Exam tips

Asked in 2025 Q6(ii) (5 marks): depressions of $3 5^{\circ}$ and $1 4^{\circ}$ from the top and the mid-point of a building 100 m from a tower; the expected working used $100 tan 3 5^{\circ}$ and $100 tan 1 4^{\circ}$ with table values rounded to two decimals, as the question instructed.
Write the reason 'alternate angles' when you move the depression to the ground; it shows the examiner where the triangle comes from.
For two objects, state 'same side' or 'opposite sides' in your working before adding or subtracting.
A boat moving away has a decreasing depression; a car approaching has an increasing one.

Practice questions on Angle of Depression

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The angle of depression of a point $B$ from a point $A$ above it is $4 0^{\circ}$ . The angle of elevation of $A$ from $B$ is:
(a) $4 0^{\circ}$
(b) $5 0^{\circ}$
(c) $9 0^{\circ}$
(d) $14 0^{\circ}$
Hint
Alternate angles.
Answer
(a)
Q2 · 2 marks · Short answerBoard level
From the top of a building 35 m high, the angle of depression of a bus stop on the level road is $6 0^{\circ}$ . Find the distance of the bus stop from the foot of the building, to 2 decimal places.
Hint
$d = \frac{35}{t a n 6 0 ^{\circ}}$ .
Answer
$\frac{35}{3} \approx 20.21$ m
Q3 · 1 mark · Assertion-reasonHarder board level
Assertion (A): If the angle of depression of a car from the top of a tower 30 m high is $3 0^{\circ}$ , the car is $103$ m from the foot of the tower.
Reason (R): $tan 3 0^{\circ} = \frac{1}{3}$ .
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).
Hint
$d = 30 \div \frac{1}{3}$ .
Answer
(b); the distance is $303$ m.
Q4 · 3 marks · Short answerHarder board level
From a bridge 12 m above a river, the angles of depression of two buoys, one upstream and one downstream in line with the point below the observer, are $2 6^{\circ}$ and $4 1^{\circ}$ . Find the distance between the buoys, to 2 decimal places. ( $tan 2 6^{\circ} = 0.4877$ , $tan 4 1^{\circ} = 0.8693$ )
Hint
Opposite sides: add.
Answer
$\frac{12}{0.4877} + \frac{12}{0.8693} \approx 24.605 + 13.804 \approx 38.41$ m
Q5 · 3 marks · Short answerHarder board level
From the top of a tower 65 m high, the angles of depression of two cars on the same side, in line with its foot, are $5 4^{\circ}$ and $3 1^{\circ}$ . Find the distance between the cars, to the nearest metre. ( $tan 5 4^{\circ} = 1.3764$ , $tan 3 1^{\circ} = 0.6009$ )
Hint
Same side: subtract.
Answer
$\frac{65}{0.6009} - \frac{65}{1.3764} \approx 108.171 - 47.225 \approx 60.95$ , so 61 m
Q6 · 4 marks · Short answerHarder board level
A drone hovers 90 m above a straight road. A car driving towards the point below the drone is seen at an angle of depression of $2 7^{\circ}$ , and 10 seconds later at $5 8^{\circ}$ . Find the distance covered, to 1 decimal place, and the car's speed in km/h, to 1 decimal place. ( $tan 2 7^{\circ} = 0.5095$ , $tan 5 8^{\circ} = 1.6003$ )
Hint
Approaching, so the angle grows; subtract the distances.
Answer
$\frac{90}{0.5095} - \frac{90}{1.6003} \approx 176.644 - 56.239 \approx 120.4$ m; speed $\approx 12.04$ m/s $\approx 43.3$ km/h.
Q7 · 3 marks · Short answerHOTS / competency
From the top of a tower of height $h$ m, the angle of depression of a boat is $θ$ with $tan θ = \frac{3}{4}$ . The boat moves 21 m straight towards the tower, and the angle of depression becomes $ϕ$ with $tan ϕ = \frac{4}{3}$ . Find $h$ .
Hint
Write both distances in terms of $h$ .
Answer
$d_{1} = \frac{4 h}{3}$ and $d_{2} = \frac{3 h}{4}$ , so $\frac{4 h}{3} - \frac{3 h}{4} = \frac{7 h}{12} = 21$ and $h = 36$ m. (Check: 48 m − 27 m = 21 m.)

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.The angle of depression is measured from?Show answer
The horizontal through the observer's eye, downwards.
2.Why does it equal the angle of elevation at the object?Show answer
Alternate angles: the horizontal at the eye is parallel to the ground.
3.Height $h$ , depression $θ$ : ground distance?Show answer
$h cot θ = \frac{h}{t a n θ}$
4.Two boats on the same side: which has the larger depression?Show answer
The nearer one.
5.m/s to km/h?Show answer
Multiply by 3.6.

Angle of Depression: frequently asked questions

What is the angle of depression?

It is the angle between the horizontal line through the observer's eye and the line of sight to an object below the observer, measured downwards from the horizontal.

Is the angle of depression equal to the angle of elevation?

Yes. The angle of depression of B seen from A equals the angle of elevation of A seen from B, because they are alternate angles between the parallel horizontal lines.

How do I find the distance of a boat from a cliff using the angle of depression?

Divide the height of the cliff by the tangent of the angle of depression: d = h ÷ tan θ. Use the four-figure table value and round only the final answer.