Heights and Distance

Q: How do you find the height of a tower from two observations on the same line?

If the elevation is alpha from the farther point, beta from the nearer one and the points are d apart, then h = d divided by (cot alpha minus cot beta). For observers on opposite sides the cotangents add instead, so h = d divided by (cot alpha plus cot beta).

Q: What is the angle of elevation of the sun when a pole's shadow equals its height?

It is 45 degrees, because tan theta = height divided by shadow = 1. If the shadow is root 3 times the height the elevation is 30 degrees, and if the height is root 3 times the shadow it is 60 degrees.

MathsSolution Of Triangles, Heights And DistancesFor JEE aspirants

Heights and distances is the branch of trigonometry that measures things that cannot be reached: the height of a tower, the width of a river, the speed of a boat. Every question turns into a right triangle in which one side and one angle are known, and the unknown side follows from $tan θ = \frac{opposite}{adjacent}$ . This page covers angles of elevation and depression, the standard configurations of heights and distances, slopes, lakes, three-dimensional cases, and ten fully solved problems.

Syllabus note. NTA removed Heights and Distances from the JEE Main mathematics syllabus in the 2024 revision, and the JEE Advanced syllabus does not list it either. The topic is still part of the Class 10 CBSE course (Some Applications of Trigonometry), of several state-board and other entrance syllabi, and the same reasoning appears inside solution-of-triangles questions. It is kept here in full. Check the current NTA syllabus before you plan revision time for it.

Key Formulas - Quick Reference

Basic model: height $=$ distance $\times tan θ$ , so $h = d tan θ$ and $d = h cot θ$
Angle of depression of $P$ from $O$ = angle of elevation of $O$ from $P$
Two points on the same side: $h = \frac{d}{cot α - cot β}$ ( $α$ from the farther point)
Two points on opposite sides: $d = h (cot α + cot β)$
Pole of height $x$ on a tower of height $H$ , seen from distance $d$ : $H = d tan β$ , $H + x = d tan α$
Tower seen from a building of height $H$ (elevation $α$ of the top, depression $β$ of the foot): height $= H (1 + tan α cot β)$
Cloud over a lake, observer $H$ above the water: $h = \frac{H ( tan β + tan α )}{tan β - tan α}$
Two points in perpendicular directions: $d^{2} = h^{2} (cot^{2} α + cot^{2} β)$
No right angle available: use the sine rule $\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C}$ in the slant triangle
Speed of a moving object $= \frac{horizontal distance covered}{time taken}$

1. Angle of Elevation and Angle of Depression

Let $O$ be the observer's eye and $O X$ the horizontal line through it.

Angle of elevation: if the object $P$ is above eye level, the angle $∠ P O X$ between the horizontal and the line of sight $O P$ is the angle of elevation of $P$ from $O$ .
Angle of depression: if $P$ is below eye level, the angle measured downward from the horizontal to $O P$ is the angle of depression of $P$ from $O$ .

Figure 1: Both angles are measured from the horizontal line through the eye. The angle of depression of

P

from

O

equals the angle of elevation of

O

from

P

(alternate angles).

Both angles are measured from the horizontal, never from the vertical.
The angle of depression of $P$ from $O$ equals the angle of elevation of $O$ from $P$ , because the two horizontals are parallel and the angles are alternate angles. This is what lets you move the marked angle into the triangle you are solving.
Unless the question gives the observer's height, treat the eye as being on the ground. If a height of the observer is given, subtract it first and add it back at the end.

2. The Basic Right-Triangle Model

Every heights and distances problem reduces to a right triangle whose vertical side is the height $h$ and whose horizontal side is the distance $d$ :

tan θ = \frac{h}{d} \Rightarrow h = d tan θ, d = h cot θ

The sine and cosine come in only when the slant distance (the line of sight itself) is involved: $sin θ = \frac{h}{line of sight}$ .

$θ$	$sin θ$	$cos θ$	$tan θ$	$cot θ$
$3 0^{\circ}$	$\frac{1}{2}$	$\frac{3}{2}$	$\frac{1}{3}$	$3$
$4 5^{\circ}$	$\frac{1}{2}$	$\frac{1}{2}$	$1$	$1$
$6 0^{\circ}$	$\frac{3}{2}$	$\frac{1}{2}$	$3$	$\frac{1}{3}$
$1 5^{\circ}$	$\frac{3 - 1}{2 2}$	$\frac{3 + 1}{2 2}$	$2 - 3$	$2 + 3$
$7 5^{\circ}$	$\frac{3 + 1}{2 2}$	$\frac{3 - 1}{2 2}$	$2 + 3$	$2 - 3$

A method that works for every question:

Draw the vertical object and the horizontal ground, and mark the observation points.
Put every given angle at the correct vertex, using alternate angles for depressions.
Name the unknown height $h$ and write each horizontal distance as $h cot θ$ .
Use the one given length to form a single equation, then solve for $h$ .
Rationalise the surd and, if the question asks, convert units (metres per second to kilometres per hour by multiplying by $\frac{18}{5}$ ).

Exam Trick

Write distances as $h cot θ$ , not heights as $d tan θ$ . The unknown is almost always the height, and the cotangent form keeps $h$ as a common factor, so it cancels or factors out in one step.

3. Two Observations from the Same Straight Line

3.1 Both points on the same side

A tower $P Q$ of height $h$ is seen from $A$ at an angle $α$ and from a nearer point $B$ at an angle $β$ , with $A B = d$ (Figure 2). Then $A Q = h cot α$ , $B Q = h cot β$ and

d = h (cot α - cot β) \Rightarrow h = \frac{d}{cot α - cot β}

Figure 2:

A Q = h cot α

and

B Q = h cot β

, so

d = A B = h (cot α - cot β)

and

h = \frac{d}{cot α - cot β}

3.2 The two points on opposite sides

If the observers stand on opposite sides of the tower and the distance between them is $d$ , the two foot distances add instead of subtracting:

d = h (cot α + cot β) \Rightarrow h = \frac{d}{cot α + cot β}

The same pair of formulas answers the reverse question: given the height, they give the distance between the observation points.

Worth Knowing

If the angles are complementary, that is $β = 9 0^{\circ} - α$ , then $cot α cot β = 1$ and the two foot distances $d_{1} = h cot α$ , $d_{2} = h cot β$ satisfy $h^{2} = d_{1} d_{2}$ . A tower seen at complementary angles from two points has height equal to the geometric mean of the two distances.

4. Flagstaffs, Poles and Buildings

4.1 A pole standing on a tower

Let a pole of height $x$ stand on a tower of height $H$ , and let both be seen from a point at distance $d$ on the ground. If the elevation of the top of the tower is $β$ and that of the top of the pole is $α$ , then

H = d tan β, H + x = d tan α \Rightarrow x = d (tan α - tan β)

Knowing any two of $H$ , $x$ , $d$ gives the third.

4.2 A tower seen from the top of a building

From the top $P$ of a building of height $H$ , let the angle of elevation of the top of a tower be $α$ and the angle of depression of its foot be $β$ (Figure 3). The horizontal through $P$ meets the tower at the height $H$ , so

d = H cot β, height of the tower = H + d tan α = H (1 + tan α cot β)

Figure 3: From the top of a building of height

H

d = H cot β

and the tower height is

H + d tan α = H (1 + tan α cot β)

5. A Cloud Above a Lake and Its Reflection

A lake surface acts as a mirror: a cloud at height $h$ above the water has its image at depth $h$ below the water. Let the observer's eye $E$ be $H$ above the lake, the angle of elevation of the cloud be $α$ and the angle of depression of the reflection be $β$ (Figure 4). With $d$ as the horizontal distance,

tan α = \frac{h - H}{d}, tan β = \frac{h + H}{d}

Figure 4: The reflection lies as far below the surface as the cloud is above it, so

tan α = \frac{h - H}{d}

and

tan β = \frac{h + H}{d}

Dividing one by the other removes $d$ :

\frac{h - H}{h + H} = \frac{tan α}{tan β} \Rightarrow h = \frac{H ( tan β + tan α )}{tan β - tan α}

Note that the observer's height $H$ is measured from the water surface, and that $β > α$ always, because the reflection is farther from the eye level than the cloud.

6. Moving Objects: Finding a Speed

When a boat, car or aeroplane moves while the observer stays fixed, the angle changes with time. The object's height above the ground stays the same, so only the horizontal distance changes:

Write the two horizontal distances as $h cot α$ and $h cot β$ .
Their difference is the distance covered: $h (cot α - cot β)$ .
Divide by the time taken to get the speed, then convert the units the question asks for.

The same idea gives the remaining time: if the object keeps the same speed, time $=$ remaining distance $\div$ speed.

7. Observations Taken on a Slope

When the observer walks up an incline, the two observation points are no longer on one horizontal line, so there is no ready-made right triangle. Use the slant triangle formed by the two observation points and the object, and solve it with the sine rule (Figure 5).

Every elevation is still measured from the horizontal through that point, so the angle between the line of sight and the slope is the difference of the two given angles.
At the upper point the angle inside the triangle is $18 0^{\circ}$ minus that difference.
The third angle follows from the angle sum, and the sine rule gives the slant distance to the object.
Multiply by the sine of the first elevation to get the vertical height above the starting point.

Figure 5: In

△ A B D

∠ A B D = 4 7^{\circ} - 3 2^{\circ} = 1 5^{\circ}

and

∠ A D B = 18 0^{\circ} - (7 7^{\circ} - 3 2^{\circ}) = 13 5^{\circ}

, so the sine rule gives

A B

8. Three-Dimensional Problems

In these questions the two observation points are not in line with the foot of the tower; directions such as due south, due east or bearings are given instead. The tower stands at $O$ , and the foot distances $O A$ and $O B$ lie in the horizontal plane (Figure 6).

Work in the vertical triangles first: $O A = h cot α$ and $O B = h cot β$ .
Then work in the horizontal plane. For perpendicular directions, Pythagoras gives $A B^{2} = O A^{2} + O B^{2}$ , that is $d^{2} = h^{2} (cot^{2} α + cot^{2} β)$ .
If the directions are not perpendicular but make a known angle $γ$ at $O$ , use the cosine rule in the horizontal triangle:
$d^{2} = h^{2} (cot^{2} α + cot^{2} β - 2 cot α cot β cos γ)$

Figure 6:

O A = h cot α

O B = h cot β

and

∠ A O B = 9 0^{\circ}

, so

d^{2} = h^{2} (cot^{2} α + cot^{2} β)

9. Solved Examples

Solved Example 1

A boat is being rowed away from a cliff

150

m high. At the top of the cliff the angle of depression of the boat changes from

6 0^{\circ}

4 5^{\circ}

2

minutes. Find the speed of the boat.

Solution:

Let $P Q$ be the cliff with $P Q = 150$ m, and let $A$ and $B$ be the two positions of the boat, $A$ nearer the cliff.
Depressions equal elevations from the boat, so the elevation of the top is $6 0^{\circ}$ at $A$ and $4 5^{\circ}$ at $B$ .
$A Q = 150 cot 6 0^{\circ} = \frac{150}{3}$ and $B Q = 150 cot 4 5^{\circ} = 150$ .
Distance rowed: $A B = 150 - \frac{150}{3} = 503 (3 - 1) = 150 - 503 \approx 63.4$ m in $2$ minutes.
Speed $= \frac{150 - 50 3}{2}$ metres per minute. Multiplying by $60$ : speed $= 30 (150 - 503) = 1500 (3 - 3)$ metres per hour.

Answer: $1500 (3 - 3) \approx 1902$ metres per hour, about $31.7$ m/min.

Solved Example 2

The angle of elevation of the top of a rock from a point

B

4 7^{\circ}

. After walking

1000

m from

B

towards the rock up a slope inclined at

3 2^{\circ}

to the horizontal, the elevation of the top is

7 7^{\circ}

. Find the vertical height of the rock above

B

, given

sin 4 7^{\circ} = 0.73

Solution:

Let $A$ be the top of the rock and $D$ the point reached after the walk (Figure 5). In $△ A B D$ , $∠ A B D = 4 7^{\circ} - 3 2^{\circ} = 1 5^{\circ}$ .
At $D$ the elevation is measured from the horizontal, so the angle between $D A$ and the slope produced is $7 7^{\circ} - 3 2^{\circ} = 4 5^{\circ}$ , giving $∠ A D B = 18 0^{\circ} - 4 5^{\circ} = 13 5^{\circ}$ .
Hence $∠ B A D = 18 0^{\circ} - 1 5^{\circ} - 13 5^{\circ} = 3 0^{\circ}$ .
Sine rule in $△ A B D$ : $\frac{A B}{sin 13 5 ^{\circ}} = \frac{1000}{sin 3 0 ^{\circ}}$ , so $A B = 2000 sin 13 5^{\circ} = 10002$ m.
The height above $B$ is $A C = A B sin 4 7^{\circ} = 10002 \times 0.73 = 7302$ m.

Answer: $7302 \approx 1032$ m above the first point of observation.

Solved Example 3

The angle of elevation of the top of a tower from a point

30

m from its foot on level ground is

6 0^{\circ}

. Find the height of the tower.

Solution:

$h = d tan θ = 30 tan 6 0^{\circ}$ .
$tan 6 0^{\circ} = 3$ .

Answer: $h = 303 \approx 51.96$ m

Solved Example 4

The angle of elevation of the top of a tower from a point

A

on level ground is

3 0^{\circ}

. On walking

40

m towards the tower the elevation becomes

6 0^{\circ}

. Find the height of the tower.

Solution:

With $α = 3 0^{\circ}$ at the farther point and $β = 6 0^{\circ}$ at the nearer one, $d = 40$ m.
$h = \frac{d}{cot α - cot β} = \frac{40}{3 - \frac{1}{3}}$ .
$3 - \frac{1}{3} = \frac{2}{3}$ , so $h = \frac{40 3}{2}$ .

Answer: $h = 203 \approx 34.64$ m

Solved Example 5

Two people stand on opposite sides of a tower and observe its top at angles of elevation

3 0^{\circ}

and

4 5^{\circ}

. They are

100

m apart. Find the height of the tower.

Solution:

The foot distances are $h cot 3 0^{\circ} = h 3$ and $h cot 4 5^{\circ} = h$ , and they add up to $100$ m.
$h (3 + 1) = 100$ , so $h = \frac{100}{3 + 1}$ .
Rationalise: $h = \frac{100 ( 3 - 1 )}{2} = 50 (3 - 1)$ .

Answer: $h = 50 (3 - 1) \approx 36.6$ m

Solved Example 6

A flagstaff stands on the top of a tower

10

m high. At a point on the ground the angles of elevation of the bottom and the top of the flagstaff are

3 0^{\circ}

and

4 5^{\circ}

. Find the length of the flagstaff.

Solution:

Distance of the point from the foot: $d = 10 cot 3 0^{\circ} = 103$ m.
Height of the top of the flagstaff: $d tan 4 5^{\circ} = 103$ m.
Length of the flagstaff $= 103 - 10$ .

Answer: $10 (3 - 1) \approx 7.32$ m

Solved Example 7

From the top of a building

15

m high, the angle of elevation of the top of a tower is

3 0^{\circ}

and the angle of depression of its foot is

4 5^{\circ}

. Find the height of the tower.

Solution:

The horizontal distance is $d = H cot β = 15 cot 4 5^{\circ} = 15$ m.
Above the level of the building top the tower rises $d tan 3 0^{\circ} = \frac{15}{3} = 53$ m.
Total height $= 15 + 53$ .

Answer: $15 + 53 \approx 23.66$ m

Solved Example 8

From a point

25

m above a lake, the angle of elevation of a cloud is

1 5^{\circ}

and the angle of depression of its reflection in the lake is

4 5^{\circ}

. Find the height of the cloud above the lake.

Solution:

Using the standard result with $H = 25$ , $α = 1 5^{\circ}$ , $β = 4 5^{\circ}$ :
$h = \frac{H ( tan β + tan α )}{tan β - tan α} = 25 \cdot \frac{1 + ( 2 - 3 )}{1 - ( 2 - 3 )}$
$\frac{3 - 3}{3 - 1} = \frac{3 ( 3 - 1 )}{3 - 1} = 3$ .

Answer: $h = 253 \approx 43.3$ m above the lake.

Solved Example 9

A tower stands on level ground. From a point due south of it the angle of elevation of the top is

3 0^{\circ}

, and from a point due east of it the elevation is

4 5^{\circ}

. The two points are

100

m apart. Find the height of the tower.

Solution:

Let $h$ be the height and $O$ the foot. Then $O A = h cot 3 0^{\circ} = h 3$ and $O B = h cot 4 5^{\circ} = h$ .
South and east are perpendicular directions, so $∠ A O B = 9 0^{\circ}$ and $A B^{2} = O A^{2} + O B^{2}$ .
$10 0^{2} = 3 h^{2} + h^{2} = 4 h^{2}$ , so $h = 50$ .

Answer: $h = 50$ m

Solved Example 10

A man sees the top of a hill at an angle of elevation of

4 5^{\circ}

. He walks

500

m towards it along a road inclined at

1 5^{\circ}

to the horizontal, and the elevation becomes

7 5^{\circ}

. Find the height of the hill above his first position.

Solution:

Let $A$ be the hill top, $B$ the first point and $D$ the point after the walk, with $B D = 500$ m along the slope.
$∠ A B D = 4 5^{\circ} - 1 5^{\circ} = 3 0^{\circ}$ , and $∠ A D B = 18 0^{\circ} - (7 5^{\circ} - 1 5^{\circ}) = 12 0^{\circ}$ .
So $∠ B A D = 3 0^{\circ}$ , and the sine rule gives $A B = \frac{500 sin 12 0 ^{\circ}}{sin 3 0 ^{\circ}} = 5003$ m.
Height above $B$ is $A B sin 4 5^{\circ} = \frac{500 3}{2} = 2506$ m.

Answer: $2506 \approx 612.4$ m

Practice Questions

A ladder leans against a wall making $6 0^{\circ}$ with the ground, and its foot is $2.5$ m from the wall. Find the length of the ladder.Answer: $5$ m (length $= 2.5 sec 6 0^{\circ}$ )
The shadow of a tower lengthens by $40$ m when the elevation of the sun falls from $4 5^{\circ}$ to $3 0^{\circ}$ . Find the height of the tower.Answer: $20 (3 + 1) \approx 54.64$ m
The angle of elevation of the top of a tower from a point is $3 0^{\circ}$ ; after advancing $150$ m towards it the elevation is $6 0^{\circ}$ . Find the height.Answer: $753 \approx 129.9$ m
From the top of a lighthouse $60$ m high the angles of depression of two ships on the same side are $3 0^{\circ}$ and $4 5^{\circ}$ . Find the distance between the ships.Answer: $60 (3 - 1) \approx 43.9$ m
Two men on either side of a tower $30$ m high observe its top at $3 0^{\circ}$ and $4 5^{\circ}$ . Find the distance between the men.Answer: $30 (3 + 1) \approx 81.96$ m
A vertical pole stands on a tower $50$ m high. From a point on the ground the elevations of the bottom and the top of the pole are $4 5^{\circ}$ and $6 0^{\circ}$ . Find the length of the pole.Answer: $50 (3 - 1) \approx 36.6$ m
From a point $60$ m above a lake the elevation of a cloud is $3 0^{\circ}$ and the depression of its reflection is $6 0^{\circ}$ . Find the height of the cloud above the lake.Answer: $120$ m
A tower is seen from a point due south of it at an elevation of $4 5^{\circ}$ and from a point due west of it at $6 0^{\circ}$ . The two points are $10$ m apart. Find the height.Answer: $53 \approx 8.66$ m
An aeroplane flying horizontally at $3000$ m is seen at an elevation of $6 0^{\circ}$ ; after $15$ seconds the elevation is $3 0^{\circ}$ . Find its speed in km/h.Answer: $20003$ m in $15$ s, that is about $831$ km/h
A tower stands on the bank of a river. From the opposite bank its elevation is $6 0^{\circ}$ ; from a point $20$ m further back it is $3 0^{\circ}$ . Find the height of the tower and the width of the river.Answer: Height $103 \approx 17.32$ m, width $10$ m

Common Mistakes to Avoid

Watch out

Measuring an angle of elevation or depression from the vertical. Both are measured from the horizontal line through the eye.
Marking the angle of depression inside the triangle at the object. Shift it to the correct vertex first, using alternate angles.
Ignoring the observer's height when the question gives it, or forgetting to add it back at the end.
Using $cot β - cot α$ instead of $cot α - cot β$ . The farther point has the smaller angle and the larger cotangent, so its cotangent comes first.
Treating the walking distance in a slope problem as horizontal. On an incline the two observation points are at different heights, so use the sine rule.
In cloud problems, measuring $H$ from the ground instead of from the water surface, or forgetting that the reflection is $h$ below the surface, not $h - H$ .
In three-dimensional problems, taking the distance between the two observation points as the difference of the foot distances. Use Pythagoras or the cosine rule in the horizontal plane.
Adding the height of a pole to the height of a tower when the question gives only the elevation of the top of the pole; the pole length is $d (tan α - tan β)$ .
Reporting a speed in the wrong unit. Multiply metres per second by $\frac{18}{5}$ to get kilometres per hour.

Frequently Asked Questions

What is the angle of elevation?

The angle of elevation of an object is the angle between the horizontal line through the observer's eye and the line of sight to the object, when the object is above eye level. If the horizontal distance is $d$ , the height above eye level is $d tan θ$ .

What is the difference between angle of elevation and angle of depression?

Both are measured from the horizontal through the eye. Elevation is used when the object is above eye level and depression when it is below. The angle of depression of $P$ from $O$ equals the angle of elevation of $O$ from $P$ , because the two horizontal lines are parallel.

How do you find the height of a tower from two observations on the same line?

If the elevation is $α$ from the farther point, $β$ from the nearer one and the points are $d$ apart, then $h = \frac{d}{cot α - cot β}$ . For observers on opposite sides the cotangents add instead: $h = \frac{d}{cot α + cot β}$ .

How do you solve cloud and reflection problems on a lake?

The reflection lies as far below the water as the cloud is above it. With the eye $H$ above the lake, elevation $α$ of the cloud and depression $β$ of the reflection, $tan α = \frac{h - H}{d}$ and $tan β = \frac{h + H}{d}$ , so $h = \frac{H ( tan β + tan α )}{tan β - tan α}$ .

What do you do when the observer walks up a slope?

The two observation points are then at different heights, so no right triangle joins them. Work in the slant triangle: each elevation is still measured from the horizontal, so the angle at the lower point is the difference of the two angles, and the sine rule gives the slant distance to the object.

How do you find the speed of a boat or an aeroplane from changing angles?

The height stays fixed, so only the horizontal distance changes. Write the two distances as $h cot α$ and $h cot β$ , subtract to get the distance covered, and divide by the time. Multiply metres per second by $\frac{18}{5}$ for kilometres per hour.

What is the angle of elevation of the sun when a pole's shadow equals its height?

It is $4 5^{\circ}$ , because $tan θ = \frac{height}{shadow} = 1$ . If the shadow is $3$ times the height the elevation is $3 0^{\circ}$ , and if the height is $3$ times the shadow it is $6 0^{\circ}$ .

Is Heights and Distances part of the JEE syllabus?

NTA removed it from the JEE Main mathematics syllabus in the 2024 revision and the JEE Advanced syllabus does not list it. It remains in the Class 10 CBSE course and in several other entrance syllabi, and the same right-triangle reasoning is used in solution-of-triangles questions, so it is still worth knowing.

Previous year questions on Heights and Distance

2 questions from past papers, each with a step-by-step solution.

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