Heights and Distances

high importance~2 Q in Tier 120 formulas⚡ 11 shortcuts5 subtopics

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Angles of elevation and depression

⏱ 5 min read🧩 5 question types🎯 12 practice Q

The idea in one minute

Elevation means looking up at an angle. Depression means looking down. Both angles sit between the horizontal line through the observer's eye and the line of sight.

Each question hides one right triangle. The tower is the side opposite the angle. The ground distance is the side next to it. So one ratio, $\tan\theta$ , solves most of them.

First turn every depression into an elevation. Then solve the triangle.

Looking up: the angle of elevation

Stand on flat ground and look straight ahead. Now raise your eyes to the top of a tower.

The angle your line of sight turns up from the horizontal is the angle of elevation.

The horizontal ray and the tower make a right angle. So the figure is a right triangle:

Tower = the side opposite the angle.
Ground distance = the side next to the angle.
Eye level = the height of the observer's eye above the ground.

Rule: $\tan\theta = \dfrac{\text{height above eye level}}{\text{horizontal distance}}$

A tower top is seen at $45^\circ$ from a point on the ground. $\tan 45^\circ = 1$ , so the tower is as tall as the point is far: tower 20 m, distance 20 m.

Looking down: the angle of depression

Now stand on the tower and look at a car on the road.

The angle your line of sight turns down from your horizontal line is the angle of depression.

Rule: Depression from the top equals elevation from the bottom. The two horizontal lines are parallel, so the two angles are equal (alternate angles).

This one line converts every depression question into the elevation question you already know.

Example: From the top of a 60 m tower, the angle of depression of a car is $45^\circ$ . Then the car sees the tower top at $45^\circ$ , so the car is $60 \times \cot 45^\circ = 60$ m from the foot.

Solve the triangle with tan

Write the ratio before touching any number.

h = d \times \tan\theta \qquad d = h \times \cot\theta

Here $h$ is the height above the observer's eye and $d$ is the horizontal distance.

From a point 30 m from the foot of a pole, its top is at $30^\circ$ . Height $= 30 \times \tan 30^\circ = 30 \times \dfrac{1}{\sqrt{3}} = 10\sqrt{3}$ m.

Tip: Keep $\sqrt{3}$ as $\sqrt{3}$ until the last step. Only use $1.73$ when the question says "approximately".

When the slanting length is given

A kite thread, a wire or a ladder runs along the line of sight. That slanting line is the hypotenuse.

Height reached: $h = L\sin\theta$ .
Horizontal distance: $d = L\cos\theta$ .

A kite flies on a 100 m thread at $30^\circ$ . Height $= 100 \times \dfrac{1}{2} = 50$ m. Horizontal distance $= 100 \times \dfrac{\sqrt{3}}{2} = 50\sqrt{3}$ m.

Watch: The thread is the slanting side, not the height. Options that treat the thread as the height are always present.

Shadows: similar triangles

The sun makes the same angle with every vertical object at the same time. So the two triangles are similar, and you can compare without trigonometry:

\dfrac{\text{tower height}}{\text{tower shadow}} = \dfrac{\text{stick height}}{\text{stick shadow}}

A 1.5 m stick casts a 2.5 m shadow. A tower casts a 75 m shadow at the same moment. Tower $= 75 \times \dfrac{1.5}{2.5} = 45$ m.

Eye level matters

Every angle starts at the observer's eye, not the ground.

An observer 1.5 m tall sees a tower top at $45^\circ$ from 20 m away. Height above eye level $= 20$ m. Tower $= 20 + 1.5 = 21.5$ m.

Watch: When the question gives a man's height, add it at the end. When it says "from a point on the ground", the eye is at ground level and nothing is added.

Question types you will see

Each type: how to recognise it, the method step by step, and one question to try.

Type 1very common2 practice Q

One angle at eye level

How to spot it:

A single angle of elevation from a point on the ground; a height or a distance is asked.

h = d\tan\theta

Method

Draw the right triangle: angle at the observer, tower opposite.
Mark what is known: height, distance or angle.
Write $\tan\theta = h/d$ and solve for the missing side.

Why it works:

One angle and one side fix a right triangle completely.

Try this

From a point on the ground 30 m from the foot of a tower, the angle of elevation of the top is $30^\circ$ . The height of the tower is:

Show solution

$\tan 30^\circ = \dfrac{h}{30}$ .
$h = 30 \times \dfrac{1}{\sqrt{3}} = \dfrac{30}{\sqrt{3}}$ .
$h = 10\sqrt{3}$ m.

Answer

$10\sqrt{3}$ m

Type 2very common3 practice Q

Angle of depression from a height

How to spot it:

Someone on a tower, cliff or lighthouse looks down at a boat, car or person.

d = h\cot\theta \quad (\theta = \text{angle of depression})

Method

Convert the depression into an equal elevation at the object below.
The height of the tower is the side opposite the new angle.
Distance $= h \times \cot\theta$ .

Why it works:

The downward angle and the bottom angle are equal, so the triangle is the usual one.

Try this

From the top of a lighthouse 60 m above the sea, the angle of depression of a boat is $45^\circ$ . The horizontal distance of the boat from the foot of the lighthouse is:

Show solution

Boat sees the top at $45^\circ$ .
$\tan 45^\circ = \dfrac{60}{d} = 1$ .
$d = 60$ m.

Answer

60 m

Type 3common3 practice Q

Slanting length given (thread, wire, ladder)

How to spot it:

The length of a kite thread, wire or ladder is given with its angle, not the ground distance.

h = L\sin\theta,\qquad d = L\cos\theta

Method

Spot that the given length runs along the line of sight: it is the hypotenuse.
Height $= L \times \sin\theta$ .
Distance $= L \times \cos\theta$ .

Why it works:

In a right triangle the hypotenuse pairs with the sine (opposite) and cosine (adjacent).

Try this

A kite is flying with 100 m of thread released, and the thread makes $30^\circ$ with the ground. Taking the thread as a straight line, the height of the kite above the ground is:

Show solution

$h = 100 \times \sin 30^\circ$ .
$h = 100 \times \dfrac{1}{2} = 50$ m.

Answer

50 m

Type 4common3 practice Q

Shadow of a tower and a stick

How to spot it:

A pole or stick and its shadow are given; the shadow of a tower at the same time is asked.

\frac{H}{\text{tower shadow}} = \frac{\text{stick}}{\text{stick shadow}}

Method

Write the stick ratio: height over shadow.
Multiply the tower's shadow by that ratio.
No tangent is needed when the sun angle is not given.

Why it works:

The sun makes equal angles with both objects, so the two right triangles are similar.

Try this

A vertical stick 1.5 m long casts a shadow 2.5 m long. At the same time, a tower casts a shadow 75 m long. The height of the tower is:

Show solution

Ratio $= \dfrac{1.5}{2.5} = \dfrac{3}{5}$ .
Tower $= 75 \times \dfrac{3}{5}$ .
Tower $= 45$ m.

Answer

45 m

Type 5occasional

A cloud and its reflection in a lake

How to spot it:

An observer above water sees a cloud at one angle and its reflection at another angle.

H = h\,\frac{\tan\alpha + \tan\beta}{\tan\beta - \tan\alpha}

Method

Let $H$ be the cloud height and $h$ the eye height above the water.
Elevation $\alpha$ : $\tan\alpha = (H-h)/d$ . Reflection $\beta$ : $\tan\beta = (H+h)/d$ .
Divide the two equations and solve for $H$ .

Why it works:

The reflection behaves like an object as far below the water as the cloud is above it.

Try this

A man's eye is 20 m above a lake. The angle of elevation of a cloud is $30^\circ$ and the angle of depression of its reflection is $60^\circ$ . The height of the cloud above the lake is:

Show solution

$\dfrac{\tan 60^\circ}{\tan 30^\circ} = \dfrac{H+20}{H-20}$ .
$3 = \dfrac{H+20}{H-20}$ , so $3H - 60 = H + 20$ .
$2H = 80$ , giving $H = 40$ m.

Answer

40 m

Formula sheet

Tangent rule

\tan\theta = \frac{\text{height above eye}}{\text{horizontal distance}}

The angle sits at the observer. Height is opposite, distance is next to the angle.

Height and distance

h = d\tan\theta,\qquad d = h\cot\theta

Slanting length (thread, wire, ladder)

h = L\sin\theta,\qquad d = L\cos\theta

L is the slanting line of sight, the hypotenuse of the triangle.

Depression to elevation

\text{depression from top} = \text{elevation from bottom}

The two horizontal lines are parallel, so the angles are equal.

Shortcuts that save time

⚡ Swap depression for elevation first

Never work with a downward angle directly. Redraw it at the bottom of the tower and solve an ordinary elevation question.

Example

From the top of a 30 m tower, the angle of depression of a car is $60^\circ$ . How far is the car from the foot of the tower?

Show solution

Depression $60^\circ$ means the car sees the top at $60^\circ$ .
Distance $= 30 \times \cot 60^\circ = 30 \times \dfrac{1}{\sqrt{3}}$ .
$= \dfrac{30}{\sqrt{3}} = 10\sqrt{3}$ m.

Answer

$10\sqrt{3}$ m

⚡ Shadows without trigonometry

Same sun, same time, similar triangles. Set up the stick ratio and multiply; no tangent needed.

Example

A 2 m pole casts a 4 m shadow. How tall is a tower whose shadow is 28 m at the same time?

Show solution

Ratio $= \dfrac{2}{4} = \dfrac{1}{2}$ .
Tower $= 28 \times \dfrac{1}{2} = 14$ m.

Answer

14 m

⚡ Forty-five degrees means equal legs

Whenever the angle is $45^\circ$ , the height above the eye equals the horizontal distance. Write the equal pair without any tangent.

Example

The top of a tower is seen at $45^\circ$ from a point 28 m from its foot. What is the height of the tower (eye at ground level)?

Show solution

$\tan 45^\circ = 1$ , so height $=$ distance.
Height $= 28$ m.

Answer

28 m

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Using $\sin\theta$ with the ground angle when the tower is the opposite side.

The angle is at the observer, the tower is opposite it, so use $\tan\theta$ .

Mistake 02

Measuring the angle of depression from the vertical line downward.

Depression is measured from the horizontal line through the eye, going down.

Mistake 03

Treating the kite thread or ladder length as the height.

The thread is the hypotenuse: height $= L\sin\theta$ , distance $= L\cos\theta$ .

Mistake 04

Ignoring the observer's eye height when the question gives a man's height.

Add the eye height to the height above the eye at the end.

Mistake 05

Turning a depression into an elevation and then changing its value.

The two angles are equal. Only the position of the angle moves.

Quick revision

Read this the night before the exam.

Elevation: angle up from the horizontal at the observer's eye.
Depression: angle down. It equals the elevation from the object below.
$\tan\theta = \dfrac{\text{height above eye}}{\text{distance}}$ .
Thread, wire, ladder $= L$ : use $h = L\sin\theta$ , $d = L\cos\theta$ .
Shadows: $\dfrac{\text{height}}{\text{shadow}}$ is the same for all objects at one time.
Height above eye $+$ eye height $=$ full height.

Practice: 12 questions

Sets of 10, mixed across the question types above. Every answer has a step-by-step explanation.

Topic test · 12 questions

Suggested time 7 min · wrong answers go to your mistake notebook automatically.

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