MathsICSE Class 10 · Heights and Distances

Angles of Elevation from Two Points: Same Side and Opposite Sides (ICSE Class 10)

Q: How do I find the height of a tower from two angles of elevation?

Draw two right triangles sharing the height h. Write h = x tan β for the nearer point and h = (x + d) tan α for the farther point, eliminate x, then substitute the table values.

Q: What changes when the two points are on opposite sides?

The tower stands between the points, so the distance between them is the sum of the two horizontal distances, h cot α + h cot β, not the difference.

Q: Is a problem with two angles of elevation within the ICSE syllabus?

Yes. It uses two right-angled triangles. The CISCE syllabus excludes only cases involving more than two right-angled triangles.

Updated 15 October 2026 · Fact-checked

When one height is seen from two points, draw two right triangles that share the vertical side $h$. On the same side, $h = x\tan\beta = (x + d)\tan\alpha$, so subtract the distances; on opposite sides, add them: $AB = h\cot\alpha + h\cot\beta$. Solve with letters first, then use table values.

Both right triangles share the height h, so the distance d between the points is h cot α − h cot β; the farther point sees the smaller angle.

Illustration drawn for QuestionBuddy.

Understand Angles of Elevation from Two Points

The CISCE 2027 syllabus asks for 2-D heights-and-distances problems using trigonometric tables, and excludes cases with more than two right-angled triangles. A height seen from two points is exactly two right triangles, and it is a typical 4- or 5-mark form (2024 Q5(ii)).

Set-up. Let the vertical object be $C D$ with foot $C$ and height $h$ . The two observation points $A$ and $B$ are on level ground in line with the foot $C$ . Each point gives one right triangle with the right angle at $C$ , and the two triangles share the side $C D = h$ .

Case 1: both points on the same side. The farther point sees a smaller angle. If $B$ is nearer with angle $β$ at distance $x$ , and $A$ is $d$ further away with angle $α$ ( $α < β$ ):

$h = x tan β$ and $h = (x + d) tan α$ .
Eliminate $x$ : $h = \frac{d tan α tan β}{tan β - tan α}$ .
Or work with distances: $d = h cot α - h cot β$ .

Case 2: points on opposite sides. The object stands between the points, so the horizontal distances add: $A B = h cot α + h cot β = \frac{h}{tan α} + \frac{h}{tan β}$ .

Case 3: a moving observer or object. A person walks towards a tower, or a car approaches it, and the angle changes. This is Case 1: the distance moved is $d$ . Speed $=$ distance $\div$ time; multiply m/s by 3.6 for km/h.

Using tables. For a non-standard angle, read the four-figure value (for example $tan 3 4^{\circ} = 0.6745$ ). Dividing by $tan θ$ is the same as multiplying by $cot θ = tan (9 0^{\circ} - θ)$ , which the CISCE expected answer for 2024 Q5(ii) used: $100 cot 5 2^{\circ} = 100 tan 3 8^{\circ}$ . Keep four figures through the working and round only the final answer.

A useful special case. If the angle at the nearer point is double the angle at the farther point ( $β = 2 α$ ), the triangle formed by the two points and the top is isosceles, so the line of sight from the nearer point equals $d$ . Then $h = d sin 2 α$ .

Angles of Elevation from Two Points: formulas to remember

Same side, two equations: $h = x tan β = (x + d) tan α$; $β$ at the nearer point, $α$ at the farther point, $d$ apart.
Same side, height directly: $h = \frac{d tan α tan β}{tan β - tan α}$; Use after you have shown the two equations.
Same side, gap between the points: $d = h cot α - h cot β$; Distance moved by an observer walking in.
Opposite sides: $A B = h cot α + h cot β$; The object stands between $A$ and $B$ .
Angle doubles: $β = 2 α \Rightarrow h = d sin 2 α$; The nearer line of sight equals $d$ .

How to solve Angles of Elevation from Two Points questions

Two triangles, one shared height: decide whether the horizontal distances subtract (same side) or add (opposite sides).

1Draw the object vertically with the right angle at its foot; put both points on one horizontal line through the foot.
2Mark each angle at its point, measured from the horizontal; the farther point on the same side has the smaller angle.
3Label the unknowns: $h$ for the height and $x$ for one horizontal distance.
4Write one $tan$ equation for each triangle (opposite over adjacent).
5Eliminate $x$ (same side) or add the two distances (opposite sides) before substituting any table value.
6Substitute four-figure table values or exact surds, then round once, to the accuracy asked.

Angles of Elevation from Two Points 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.

Width of a river from one bank

Suppose you stand on a river bank opposite a tall tree on the far bank, and the angle of elevation of the treetop is $6 0^{\circ}$ . You walk 20 m straight back from the river, and the angle becomes $3 0^{\circ}$ .

How wide is the river, and how tall is the tree? (Ignore your eye height.)

Try it first, then show the solutionSolution

Let the width be $x$ and the tree height $h$ . Then $h = x tan 6 0^{\circ} = 3 x$ and $h = (x + 20) tan 3 0^{\circ} = \frac{x + 20}{3}$ .
$3 x = x + 20$ , so $x = 10$ m.
$h = 103 \approx 17.32$ m.

Answer: River 10 m wide; tree about 17.32 m tall

Memory hook: Two angles, one height: you can measure a river without crossing it.

Opposite sides of a mobile tower

Imagine two friends standing on level ground on opposite sides of a mobile tower 40 m tall, in line with its foot. One sees the top at $3 1^{\circ}$ , the other at $4 7^{\circ}$ .

How far apart are the friends, to 1 decimal place? ( $tan 3 1^{\circ} = 0.6009$ , $tan 4 7^{\circ} = 1.0724$ )

Try it first, then show the solutionSolution

Distance of the first friend $= \frac{40}{0.6009} \approx 66.567$ m.
Distance of the second friend $= \frac{40}{1.0724} \approx 37.300$ m.
Opposite sides, so add: $66.567 + 37.300 \approx 103.9$ m.

Answer: About 103.9 m apart

Memory hook: Tower in the middle: add the two distances.

Tricks to remember Angles of Elevation from Two Points

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

Same side subtract, opposite sides add
Both points on one side: the gap is the difference of the two distances. Tower between them: the gap is the sum.
Use it when: Choosing the equation for the distance between the points.
Far point, small angle
The further you stand, the flatter you look: the smaller angle always goes at the farther point.
Use it when: Placing $α$ and $β$ correctly on the diagram.
30–60 means 1 : 3
With $3 0^{\circ}$ and $6 0^{\circ}$ from the same side, the farther distance is three times the nearer one, so the gap is twice the nearer distance.
Use it when: Checking 1-mark answers quickly.
Letters first, numbers last
Eliminate $x$ while everything is still $tan α$ and $tan β$ ; put in table values only in the final line.
Use it when: Avoiding long decimal arithmetic and early rounding.

Angles of Elevation from Two Points: solved examples

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

Example 1: Distance between two points (standard angles) · 1 mark · MCQBoard level

The angle of elevation of the top of a tower from a point $P$ , 10 m from its foot, is $6 0^{\circ}$ . From a point $Q$ further away on the same line, the angle is $3 0^{\circ}$ . The distance $P Q$ is:
(a) 10 m
(b) 20 m
(c) 30 m
(d) $103$ m

Try it first, then show the solutionSolution

$h = 10 tan 6 0^{\circ} = 103$ m.
Distance of $Q = \frac{10 3}{t a n 3 0 ^{\circ}} = 103 \times 3 = 30$ m.
$P Q = 30 - 10 = 20$ m.

Answer: (b)

Example 2: Same side with 45° and 60° · 4 marks · Short answerBoard level

From two points $A$ and $B$ on level ground, 40 m apart and on the same side of a tower in line with its foot, the angles of elevation of the top are $4 5^{\circ}$ and $6 0^{\circ}$ . Find the height of the tower and the distance of $B$ (the nearer point) from its foot, to 2 decimal places.

Try it first, then show the solutionSolution

Let the height be $h$ and the distance of $B$ from the foot be $x$ . In the nearer triangle, $tan 6 0^{\circ} = \frac{h}{x}$ , so $h = 3 x$ .
In the farther triangle, $tan 4 5^{\circ} = \frac{h}{x + 40}$ , so $h = x + 40$ .
$3 x = x + 40$ , so $x = \frac{40}{3 - 1} = 20 (3 + 1) \approx 54.64$ m.
$h = x + 40 = 60 + 203 \approx 94.64$ m.

Answer: Height ≈ 94.64 m; $B$ is ≈ 54.64 m from the foot

Example 3: Opposite sides with table values · 4 marks · Short answerHarder board level

A tower 72 m high stands between two points $A$ and $B$ on level ground, in line with its foot. The angles of elevation of its top from $A$ and $B$ are $3 4^{\circ}$ and $5 7^{\circ}$ . Find $A B$ to the nearest metre. ( $tan 3 4^{\circ} = 0.6745$ , $tan 5 7^{\circ} = 1.5399$ )

Try it first, then show the solutionSolution

Let the foot be $C$ . In $△ A C D$ , $tan 3 4^{\circ} = \frac{72}{A C}$ , so $A C = \frac{72}{0.6745} \approx 106.75$ m.
In $△ B C D$ , $tan 5 7^{\circ} = \frac{72}{B C}$ , so $B C = \frac{72}{1.5399} \approx 46.76$ m.
The tower is between the points, so $A B = A C + B C \approx 153.50$ m.
To the nearest metre, $A B = 154$ m.

Answer: 154 m

Example 4: Same side with table values · 5 marks · Long answerHarder board level

The angles of elevation of the top of a chimney from two points on level ground, on the same side and in line with its foot, are $2 8^{\circ}$ and $4 1^{\circ}$ . The points are 35 m apart. Find the height of the chimney and the distance of the nearer point from its foot, to 2 decimal places. ( $tan 2 8^{\circ} = 0.5317$ , $tan 4 1^{\circ} = 0.8693$ )

Try it first, then show the solutionSolution

Let the height be $h$ and the nearer distance $x$ . Nearer point: $h = x tan 4 1^{\circ}$ . Farther point: $h = (x + 35) tan 2 8^{\circ}$ .
$x tan 4 1^{\circ} = x tan 2 8^{\circ} + 35 tan 2 8^{\circ}$ , so $x = \frac{35 t a n 2 8 ^{\circ}}{t a n 4 1 ^{\circ} - t a n 2 8 ^{\circ}}$ .
$x = \frac{35 \times 0.5317}{0.8693 - 0.5317} = \frac{18.6095}{0.3376} \approx 55.12$ m.
$h = x tan 4 1^{\circ} = 55.1229 \times 0.8693 \approx 47.92$ m (using the unrounded $x$ ).

Answer: Height ≈ 47.92 m; nearer point ≈ 55.12 m from the foot

Example 5: Moving cyclist and speed · 4 marks · Short answerHOTS / competency

A cyclist rides on a level road straight towards a tower 45 m high. The angle of elevation of the top changes from $2 5^{\circ}$ to $5 0^{\circ}$ in 6 seconds. Find the distance covered, to 2 decimal places, and the cyclist's speed in km/h, to 1 decimal place. ( $tan 2 5^{\circ} = 0.4663$ , $tan 5 0^{\circ} = 1.1918$ )

Try it first, then show the solutionSolution

First distance from the foot $= \frac{45}{t a n 2 5 ^{\circ}} = \frac{45}{0.4663} \approx 96.504$ m.
Second distance $= \frac{45}{1.1918} \approx 37.758$ m.
Distance covered $= 96.504 - 37.758 \approx 58.75$ m.
Speed $= \frac{58.746}{6} \approx 9.791$ m/s $= 9.791 \times 3.6 \approx 35.2$ km/h.

Answer: ≈ 58.75 m; ≈ 35.2 km/h

Example 6: Case study: the angle doubles · 5 marks · Case-basedHOTS / competency

Riya stands at $P$ on level ground and sees the top $T$ of a temple spire at an angle of elevation of $2 6^{\circ}$ . She walks 40 m straight towards the spire to $Q$ , and the angle becomes $5 2^{\circ}$ . Let $C$ be the foot of the spire.
(a) Find $∠ P T Q$ and explain why $QT = 40$ m.
(b) Hence find the height $C T$ and the distance $QC$ , to 2 decimal places. ( $sin 5 2^{\circ} = 0.7880$ , $cos 5 2^{\circ} = 0.6157$ )
(c) Check the height with the two-equation method, using $tan 2 6^{\circ} = 0.4877$ and $tan 5 2^{\circ} = 1.2799$ .

Try it first, then show the solutionSolution

(a) $∠ T QC = 5 2^{\circ}$ is an exterior angle of $△ P QT$ , so $∠ P T Q = 5 2^{\circ} - 2 6^{\circ} = 2 6^{\circ} = ∠ T P Q$ . The triangle is isosceles, so $QT = P Q = 40$ m.
(b) In right $△ QC T$ : $C T = 40 sin 5 2^{\circ} = 40 \times 0.7880 = 31.52$ m and $QC = 40 cos 5 2^{\circ} = 40 \times 0.6157 \approx 24.63$ m.
(c) $h = \frac{40 t a n 2 6 ^{\circ} t a n 5 2 ^{\circ}}{t a n 5 2 ^{\circ} - t a n 2 6 ^{\circ}} = \frac{40 \times 0.4877 \times 1.2799}{0.7922} \approx 31.52$ m ✓

Answer: (a) $2 6^{\circ}$ ; isosceles triangle (b) $C T$ = 31.52 m, $QC$ ≈ 24.63 m (c) ≈ 31.52 m, agreeing

Where students lose marks in Angles of Elevation from Two Points

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

Mistake: Writing tan as adjacent over opposite
With the angle at the observer, the height is opposite and the ground distance is adjacent.
Wrong working
$tan 5 2^{\circ} = \frac{A C}{100}$
Right working
$tan 5 2^{\circ} = \frac{100}{A C}$ , so $A C = 100 cot 5 2^{\circ}$
Do this: Name the opposite and adjacent sides on the diagram before writing any ratio.
Usually costs: Reported in the CISCE examiners' comments on 2024 Q5(ii); the error loses most of the question.
Mistake: Adding distances when the points are on the same side
On the same side the gap is the difference of the two distances.
Wrong working
$d = h cot α + h cot β$ for two points on one side
Right working
$d = h cot α - h cot β$
Do this: Read the question for 'same side' or 'opposite sides' and draw it.
Usually costs: The method marks for the second equation and the answer.
Mistake: Rounding a table value or an intermediate distance
Early rounding shifts the final digit.
Wrong working
$tan 5 2^{\circ} \approx 1.28$ , then a long division
Right working
Use 1.2799 (or $cot 5 2^{\circ} = tan 3 8^{\circ} = 0.7813$ ) and round only the answer
Do this: Keep four figures throughout.
Usually costs: Reported in the CISCE examiners' comments on 2024 Q5(ii): the accuracy mark.
Mistake: Not rounding as asked
'Nearest metre' needs a whole number.
Wrong working
$A B = 178.13$ m left as the answer
Right working
$A B \approx 178$ m
Do this: Underline the accuracy in the question and finish with it.
Usually costs: Reported in the CISCE examiners' comments on 2024 Q5(ii): 1 mark.

Exam tips

Asked in 2024 Q5(ii) (5 marks): a 100 m tree between two points with elevations $5 2^{\circ}$ and $4 5^{\circ}$ ; the expected answer was $100 + 100 tan 3 8^{\circ} \approx 178$ m, to the nearest metre.
The CISCE examiners suggest practising non-standard angles with tables and rounding only at the end.
Show both $tan$ equations before eliminating; each one usually carries a method mark.
With standard angles, keep surds ( $3$ ) until the last line, then use $3 \approx 1.732$ only if a decimal is asked.
For speed, convert m/s to km/h by multiplying by 3.6.

Practice questions on Angles of Elevation from Two Points

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
A pole 15 m high stands between two points on level ground, in line with its foot. The angle of elevation of its top from each point is $4 5^{\circ}$ . The distance between the points is:
(a) 15 m
(b) 30 m
(c) $153$ m
(d) $152$ m
Hint
Each distance equals the height.
Answer
(b)
Q2 · 3 marks · Short answerBoard level
The angles of elevation of the top of a tower 18 m high from two points on the same side, in line with its foot, are $3 0^{\circ}$ and $6 0^{\circ}$ . Find the distance between the points, to 2 decimal places.
Hint
Distances are $18 cot 3 0^{\circ}$ and $18 cot 6 0^{\circ}$ .
Answer
$183 - \frac{18}{3} = 183 - 63 = 123 \approx 20.78$ m
Q3 · 1 mark · Assertion-reasonHarder board level
Assertion (A): If the angles of elevation of the top of a tower from two points on the same side are $3 0^{\circ}$ and $6 0^{\circ}$ , the distance between the points is twice the distance of the nearer point from the foot.
Reason (R): $tan 6 0^{\circ} = 3 tan 3 0^{\circ}$ .
(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
If the nearer distance is $x$ , the farther one is $\frac{h}{t a n 3 0 ^{\circ}} = 3 x$ .
Answer
(c)
Q4 · 4 marks · Short answerHarder board level
Two points $A$ and $B$ on level ground are 150 m apart, with a vertical tower between them in line with its foot. The angles of elevation of its top from $A$ and $B$ are $2 9^{\circ}$ and $6 3^{\circ}$ . Find the height of the tower to 1 decimal place. ( $tan 2 9^{\circ} = 0.5543$ , $tan 6 3^{\circ} = 1.9626$ )
Hint
$h cot 2 9^{\circ} + h cot 6 3^{\circ} = 150$ .
Answer
$h (\frac{1}{0.5543} + \frac{1}{1.9626}) = 150$ , so $h \times 2.3136 \approx 150$ and $h \approx 64.8$ m.
Q5 · 4 marks · Short answerHarder board level
From two points 45 m apart on the same side of a building, in line with its foot, the angles of elevation of its top are $2 2^{\circ}$ and $3 7^{\circ}$ . Find the height of the building and the distance of the nearer point from its foot, to 2 decimal places. ( $tan 2 2^{\circ} = 0.4040$ , $tan 3 7^{\circ} = 0.7536$ )
Hint
$x = \frac{45 t a n 2 2 ^{\circ}}{t a n 3 7 ^{\circ} - t a n 2 2 ^{\circ}}$ .
Answer
$x = \frac{18.18}{0.3496} \approx 52.00$ m; $h = 52.0023 \times 0.7536 \approx 39.19$ m.
Q6 · 4 marks · Short answerHarder board level
A girl sees the top of a water tank tower at an angle of elevation of $2 7^{\circ}$ . After walking 25 m straight towards it on level ground, the angle is $4 3^{\circ}$ . Find the height of the tower and her new distance from its foot, each to 1 decimal place. ( $tan 2 7^{\circ} = 0.5095$ , $tan 4 3^{\circ} = 0.9325$ )
Hint
Same-side case with $d = 25$ .
Answer
$x = \frac{25 \times 0.5095}{0.9325 - 0.5095} = \frac{12.7375}{0.423} \approx 30.1$ m; $h = 30.112 \times 0.9325 \approx 28.1$ m.
Q7 · 3 marks · Short answerHOTS / competency
The angles of elevation of the top of a pole from two points 5 m and 20 m from its foot, on the same side and in line with it, are complementary. Prove that the height is $5 \times 20$ m, and find it.
Hint
If one angle is $θ$ , the other is $9 0^{\circ} - θ$ , and $tan (9 0^{\circ} - θ) = cot θ$ .
Answer
$tan θ = \frac{h}{5}$ (nearer, larger angle) and $cot θ = \frac{h}{20}$ . Multiplying, $1 = \frac{h ^{2}}{100}$ , so $h = 100 = 10$ 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.Same side: which point has the smaller angle?Show answer
The farther point.
2.Opposite sides: distance between the points?Show answer
$h cot α + h cot β$
3.Same side: distance between the points?Show answer
$h cot α - h cot β$ ( $α$ the smaller angle)
4. $cot 5 2^{\circ}$ from the tan table?Show answer
$tan 3 8^{\circ} = 0.7813$
5.If the angle doubles from $α$ to $2 α$ after walking $d$ , the height is?Show answer
$d sin 2 α$
6.Complementary angles from distances $a$ and $b$ give height?Show answer
$ab$

Angles of Elevation from Two Points: frequently asked questions

How do I find the height of a tower from two angles of elevation?

Draw two right triangles sharing the height h. Write h = x tan β for the nearer point and h = (x + d) tan α for the farther point, eliminate x, then substitute the table values.

What changes when the two points are on opposite sides?

The tower stands between the points, so the distance between them is the sum of the two horizontal distances, h cot α + h cot β, not the difference.

Is a problem with two angles of elevation within the ICSE syllabus?

Yes. It uses two right-angled triangles. The CISCE syllabus excludes only cases involving more than two right-angled triangles.