Heights and Distances ICSE Class 10: Angles of Elevation and Depression

What Heights and Distances is about

Weightage: CISCE publishes no chapter or unit weightage for ICSE Mathematics. Heights and Distances is in the 5. Trigonometry (b) part of the 2027 syllabus and can be asked in both Section A and Section B of the 80-mark paper.Source: CISCE ICSE Mathematics syllabus, examination year 2027

Key concepts

Angles of elevation and depression

• Angle of elevation: from the horizontal up to the line of sight (observer below the object).
• Angle of depression: from the horizontal down to the line of sight (observer above the object).

The horizontal through the observer and the ground are parallel, so the angle of depression from $A$ to $B$ equals the angle of elevation from $B$ to $A$ (alternate angles). Always measure angles from the horizontal, never from the vertical.

Worked example

From the top of a cliff, the angle of depression of a boat is $30^{\circ }$. What is the angle of elevation of the top of the cliff from the boat, and why?

Try it first, then show the solutionSolution

1. The horizontal at the top of the cliff is parallel to the sea surface.
2. The line of sight is a transversal, so the alternate angles are equal: $30^{\circ }$.

Answer: 30°, by alternate angles

Choosing the ratio

Label the right triangle with the angle, then pick the ratio that links the side you know to the side you want:

• height and horizontal distance: $\tan \theta =\frac{\text{opposite}}{\text{adjacent}}$ (the most common);
• a sloping length (ladder, kite string, wire) and the height: $\sin \theta =\frac{\text{opp}}{\text{hyp}}$;
• a sloping length and the horizontal distance: $\cos \theta =\frac{\text{adj}}{\text{hyp}}$.

Put the unknown in the numerator where possible, e.g. $h=d\tan \theta$, to avoid dividing.

Worked example

A tower stands on level ground. From a point 20 m from its foot, the angle of elevation of its top is $60^{\circ }$. Find its height.

Try it first, then show the solutionSolution

1. $\tan 60^{\circ }=\frac{h}{20}$.
2. $h=20\sqrt{3}\approx 20\times 1.732=34.64$ m.

Answer: 34.64 m

Using trigonometric tables

For non-standard angles, read the value from the four-figure tables, e.g. $\tan 50^{\circ }=1.1918$, $\tan 35^{\circ }=0.7002$, $\sin 40^{\circ }=0.6428$.

• Use the value as printed (four decimal places) unless the question says to round it (some questions say 'correct to two decimal places').
• Keep full values through the working; round only the final answer to the accuracy asked (nearest metre, 2 d.p., etc.).
• For standard angles use exact values: $\tan 30^{\circ }=\frac{1}{\sqrt{3}}$, $\tan 45^{\circ }=1$, $\tan 60^{\circ }=\sqrt{3}\approx 1.732$.

Worked example

From a point on level ground, the angle of elevation of the top of a 100 m tower is $50^{\circ }$. How far is the point from the foot of the tower? ($\tan 50^{\circ }=1.1918$)

Try it first, then show the solutionSolution

1. $\tan 50^{\circ }=\frac{100}{d}$, so $d=\frac{100}{1.1918}$.
2. $d\approx 83.91$ m.

Answer: 83.91 m

Two observation points on the same side

An object of height $h$ is seen from $A$ and $B$ on the same side, in line with its foot, at elevations $\alpha$ (farther) and $\beta$ (nearer). If $x$ is the nearer distance:
$h=x\tan \beta ,h=(x+d)\tan \alpha ,$
where $d=AB$. Solve the two equations together (eliminate $x$ or $h$).

Worked example

The angles of elevation of the top of a tower from two points 40 m apart, on the same side and in line with the foot, are $30^{\circ }$ and $60^{\circ }$. Find the height.

Try it first, then show the solutionSolution

1. Nearer point: $h=x\sqrt{3}$. Farther: $h=\frac{x+40}{\sqrt{3}}$.
2. $3x=x+40$, so $x=20$ m.
3. $h=20\sqrt{3}\approx 34.64$ m.

Answer: 34.64 m

Angles of depression from a height

From the top of a cliff, tower or building of height $h$, an object on the ground at angle of depression $\theta$ is at horizontal distance
$d=\frac{h}{\tan \theta }.$
Mark $\theta$ at the top, between the horizontal and the line of sight, then copy it to the ground angle at the object (alternate angles) to form the right triangle.

Worked example

From the top of a lighthouse 100 m high, the angle of depression of a boat is $30^{\circ }$. How far is the boat from the foot of the lighthouse?

Try it first, then show the solutionSolution

1. $\tan 30^{\circ }=\frac{100}{d}$.
2. $d=100\sqrt{3}\approx 173.21$ m.

Answer: 173.21 m

Two vertical objects

From the top of a building of height $h_1$, you see the top of a taller tower at elevation $\alpha$ and its foot at depression $\beta$. With horizontal distance $d$:

• $d=\frac{h_1}{\tan \beta }$ (from the depression of the foot);
• extra height above your eye $=d\tan \alpha$;
• tower height $=h_1+d\tan \alpha$.

Draw the horizontal from the top of the building to the tower: it splits the tower into two parts, one of which equals $h_1$.

Worked example

From the top of a 20 m building, the angle of elevation of the top of a tower is $30^{\circ }$ and the angle of depression of its foot is $45^{\circ }$. Find the height of the tower.

Try it first, then show the solutionSolution

1. $d=\frac{20}{\tan 45^{\circ }}=20$ m.
2. Extra height $=20\tan 30^{\circ }=\frac{20}{\sqrt{3}}\approx 11.55$ m.
3. Tower $=20+11.55=31.55$ m.

Answer: 31.55 m

Maths in real life: Heights and Distances

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.

Topics in this chapter

• Angles of Elevation from Two Points
• Angle of Depression
• Two Vertical Objects

Heights and Distances formula sheet

Height from distance and elevation

$h=d\tan \theta$

Right angle at the foot of the object.

Distance from height and depression

$d=\frac{h}{\tan \theta }$

Depression at the top = elevation at the bottom.

Sloping length

$h=l\sin \theta ,d=l\cos \theta$

Ladder, string, wire.

Two points on the same side

$h=x\tan \beta =(x+d)\tan \alpha$

α at the farther point.

Two points on opposite sides

$AB=\frac{h}{\tan \alpha }+\frac{h}{\tan \beta }$

Add the two horizontal distances.

Standard values

$\tan 30^{\circ }=\frac{1}{\sqrt{3}},\tan 45^{\circ }=1,\tan 60^{\circ }=\sqrt{3}$

√3 ≈ 1.732.

How the board asks Heights and Distances

The question patterns to expect, and the method that works for each.

Tricks to remember Heights and Distances

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

• Draw, label, mark the right angle

  Before any ratio: sketch, label the height h and distance d, mark the right angle and the angle from the horizontal.

  Use it when: Avoiding the wrong ratio.

• TOA for towers

  Tan = Opposite ÷ Adjacent: height over ground distance.

  Use it when: Most heights-and-distances triangles.

• Depression drops to the ground

  The angle of depression at the top equals the angle of elevation at the bottom.

  Use it when: Building the triangle from a depression.

• Round once

  Keep 4-figure table values all the way; round only the answer.

  Use it when: Getting the final accuracy right.

Heights and Distances: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section B style)

Read the situation, then answer every part. In the board paper this is a 10-mark question split into 3 + 3 + 4 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Heights and Distances

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

• ✗Mistake: Using the wrong ratio

  CISCE examiners' comments, 2024 and 2025: candidates wrote, for example, tan 52° = AC/100 and tan 14° = 100/EP, putting the adjacent side over the opposite side.

  Wrong working

  tan θ = adjacent ÷ opposite

  Right working

  tan θ = opposite ÷ adjacent, read from your labelled diagram

  ✓Do this: Mark the angle, then name 'opposite' and 'adjacent' relative to that angle before writing the ratio.

  Usually costs: Most of a 4- or 5-mark question.

• ✗Mistake: Misreading or over-rounding table values

  CISCE examiners' comments, 2024 and 2025: candidates misread tan values, or rounded tan 52° to 1.28 or 1.27 while working, and some rounded values during the working instead of at the end.

  ✓Do this: Copy the four-figure value exactly (unless the question gives a rounding rule) and round only the answer.

  Usually costs: The accuracy mark.

• ✗Mistake: Not rounding the final answer as asked

  CISCE examiners' comments, 2024: candidates did not round the final answer correctly to the nearest metre.

  ✓Do this: Underline the required accuracy ('nearest metre', '2 d.p.') and finish with it.

  Usually costs: 1 mark.

• ✗Mistake: Angle of depression marked from the vertical

  The CISCE examiners' suggestions for 2025 ask candidates to draw and label diagrams, marking the angle of depression from the horizontal.

  ✓Do this: Draw the horizontal line at the observer first; the angle of depression is below it.

  Usually costs: The whole triangle set-up.

• ✗Mistake: Long arithmetic with many decimals

  CISCE examiners' comments, 2024: substituting 1.2799 at an early stage led to long calculations with errors.

  ✓Do this: Solve algebraically first (e.g. AB = h/tan α + h/tan β), then substitute once.

  Usually costs: Method and accuracy marks.

Practice questions on Heights and Distances

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

1. Q1 · 1 mark · MCQBoard level

   The shadow of a pole is equal to its height. The angle of elevation of the sun is:
   (a) $30^{\circ }$
   (b) $45^{\circ }$
   (c) $60^{\circ }$
   (d) $90^{\circ }$

   Hint

   tan θ = 1.

   Answer

   (b)

2. Q2 · 1 mark · MCQBoard level

   From a point 30 m from the foot of a tower, the angle of elevation of its top is $60^{\circ }$. The height of the tower is:
   (a) $10\sqrt{3}$ m
   (b) 30 m
   (c) $30\sqrt{3}$ m
   (d) 60 m

   Hint

   h = 30 tan 60°.

   Answer

   (c)

3. Q3 · 1 mark · MCQHarder board level

   As an observer walks along level ground straight towards a tower, the angle of elevation of its top:
   (a) decreases
   (b) increases
   (c) stays the same
   (d) first increases then decreases

   Hint

   tan θ = h/d, and d falls.

   Answer

   (b)

4. Q4 · 1 mark · Assertion-reasonBoard level

   Assertion (A): If the height of a pole and the length of its shadow are in the ratio $\sqrt{3}:1$, the sun's angle of elevation is $60^{\circ }$.
   Reason (R): $\tan 60^{\circ }=\sqrt{3}$.
   (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

   tan θ = height ÷ shadow.

   Answer

   (c)

5. Q5 · 4 marks · Short answerHarder board level

   A ladder 10 m long leans against a wall, making an angle of $62^{\circ }$ with the ground. Find, to 2 decimal places, (a) the height it reaches, (b) the distance of its foot from the wall. ($\sin 62^{\circ }=0.8829$, $\cos 62^{\circ }=0.4695$)

   Hint

   The ladder is the hypotenuse.

   Answer

   (a) $10\times 0.8829=8.83$ m (b) $10\times 0.4695=4.70$ m (4.695, rounded).

6. Q6 · 4 marks · Short answerHarder board level

   A kite flies on a string 120 m long at $40^{\circ }$ to the level ground. (a) Find its height. (b) If 30 m more string is let out at the same angle, how much higher does it rise? ($\sin 40^{\circ }=0.6428$, 2 d.p.)

   Hint

   Height is proportional to string length.

   Answer

   (a) $120\times 0.6428\approx 77.14$ m (b) $30\times 0.6428\approx 19.28$ m higher.

7. Q7 · 3 marks · Long answerBoard level

   From the top of a lighthouse 75 m high, the angle of depression of a ship is $30^{\circ }$. Find the distance of the ship from the foot of the lighthouse, to 1 decimal place.

   Hint

   d = 75 ÷ tan 30°.

   Answer

   $d=75\sqrt{3}\approx 129.9$ m

8. Q8 · 4 marks · Long answerHarder board level

   From the top of a building 15 m high, the angle of elevation of the top of a tower is $40^{\circ }$ and the angle of depression of its foot is $30^{\circ }$. Find the distance between them and the height of the tower, to 2 decimal places. ($\tan 40^{\circ }=0.8391$)

   Hint

   Depression gives the distance.

   Answer

   $d=\frac{15}{\tan 30^{\circ }}=15\sqrt{3}\approx 25.98$ m. Extra height $=25.98\times 0.8391\approx 21.80$ m. Tower $\approx 36.80$ m.

9. Q9 · 4 marks · Long answerHOTS / competency

   A man sees the top of a tower at an angle of elevation of $30^{\circ }$. After walking 20 m straight towards it, the angle becomes $60^{\circ }$. (a) Find the height of the tower. (b) How far is he now from the tower? (c) Show that the distance walked equals the length of the line of sight from his second position to the top.

   Hint

   Look for an isosceles triangle.

   Answer

   (a) $h=x\sqrt{3}=\frac{x+20}{\sqrt{3}}$ gives $x=10$, $h=10\sqrt{3}\approx 17.32$ m. (b) 10 m. (c) Line of sight $=\sqrt{10^2+300}=20$ m. In the triangle formed by the two positions and the top, the angles at the first position and at the top are both $30^{\circ }$, so it is isosceles.

10. Q10 · 4 marks · Long answerHarder board level

    A statue stands on a pedestal 4 m high. From a point on level ground, the angle of elevation of the top of the pedestal is $45^{\circ }$ and that of the top of the statue is $60^{\circ }$. Find the height of the statue to 2 decimal places.

    Hint

    The 45° gives the distance.

    Answer

    Distance $=4$ m. Top of statue $=4\sqrt{3}\approx 6.93$ m. Statue $\approx 2.93$ m.

11. Q11 · 1 mark · MCQHOTS / competency

    The angles of elevation of the top of a tower from two points 4 m and 9 m from its foot, on the same side and in line with it, are complementary. The height of the tower is:
    (a) 5 m
    (b) 6 m
    (c) 6.5 m
    (d) 13 m

    Hint

    tan θ × tan(90° − θ) = 1.

    Answer

    (b): $\frac{h}{4}\times \frac{h}{9}=1$, $h^2=36$.

12. Q12 · 5 marks · Long answerHOTS / competency

    A balloon rises vertically from a point $C$ on level ground. An observer at $A$, 200 m from $C$, sees it at an angle of elevation of $32^{\circ }$, and 30 seconds later at $50^{\circ }$. Find (a) the two heights, (b) the distance the balloon rose, (c) its average speed in m/s, to 2 decimal places. ($\tan 32^{\circ }=0.6249$, $\tan 50^{\circ }=1.1918$)

    Hint

    Each height is 200 tan θ.

    Answer

    (a) $200\times 0.6249=124.98$ m and $200\times 1.1918=238.36$ m. (b) $113.38$ m. (c) $113.38÷30\approx 3.78$ m/s.

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. 1.Angle of elevation is measured from?Show answer

   The horizontal, upwards.

2. 2.Angle of depression of B from A equals?Show answer

   The angle of elevation of A from B (alternate angles).

3. 3.Height from distance d and elevation θ?Show answer

   h = d tan θ

4. 4.Which ratio for a ladder and its height?Show answer

   sin θ = height ÷ ladder length

5. 5.tan 30°, tan 45°, tan 60°?Show answer

   1/√3, 1, √3

6. 6.When should you round?Show answer

   Only the final answer.

7. 7.Are problems with more than two right triangles in the syllabus?Show answer

   No.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Do you draw a labelled diagram every time?
• ☐Do you measure angles from the horizontal?
• ☐Do you choose tan for height and ground distance?
• ☐Do you copy table values exactly?
• ☐Do you solve algebraically before substituting?
• ☐Do you round only the final answer as asked?

Heights and Distances: frequently asked questions