Class 10 · Chapter 9 · Trigonometry unit (12 of 80 marks)

Some Applications of Trigonometry Class 10: notes and important questions

Revision notes, 10 free important questions with step-marked answers, and a four-step route from the basics to 95+ through 45 board-style questions.

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45 questions
11 multiple choice, 4 assertion–reason, 10 very short answer, 10 short answer, 6 long answer, 4 case study
About 10 hours to master

Trigonometry unit: 12 of 80 theory marks (Introduction to Trigonometry and Some Applications of Trigonometry).

NCERT textbook: Chapter 9 on ncert.nic.in (PDF)

Revision notes

Key terms

Line of sight: the line from the observer's eye to the object viewed.
Angle of elevation: the angle between the horizontal line and the line of sight when the object is above the horizontal through the eye (you raise your head).
Angle of depression: the angle between the horizontal line and the line of sight when the object is below the horizontal (you lower your head).
The angle of elevation of $Q$ from $P$ equals the angle of depression of $P$ from $Q$ . They are alternate angles between parallel horizontal lines.

Standard method

Draw a neat figure. Mark vertical objects as vertical lines, the ground as a horizontal line, and put a right angle at the foot.
Mark the given angles from the horizontal. For an angle of depression, draw the horizontal through the observer first.
Choose the ratio that links the known side with the unknown one. Usually $\tan θ = \frac{height}{horizontal distance}$ .
With two triangles, write one equation from each, then eliminate the common unknown (usually the horizontal distance).
CBSE is a no-calculator exam: leave answers in surd form (rationalise the denominator). Use $\sqrt{3} = 1.732$ or $\sqrt{2} = 1.414$ only when the question gives the value, as board papers do (“Take $\sqrt{3} = 1.73$ ”).

Useful values: $\tan 30^{\circ} = \frac{1}{\sqrt{3}}, \tan 45^{\circ} = 1, \tan 60^{\circ} = \sqrt{3}$ ; $\sin 30^{\circ} = \frac{1}{2}, \sin 45^{\circ} = \frac{1}{\sqrt{2}}, \sin 60^{\circ} = \frac{\sqrt{3}}{2}$ .

Worked example 1 (one triangle)

A 12 m ladder leans against a wall and makes $60^{\circ}$ with the ground. How high up the wall does it reach?

$\sin 60^{\circ} = \frac{h}{12} \Rightarrow h = 12 \cdot \frac{\sqrt{3}}{2} = 6 \sqrt{3}$ m.

Worked example 2 (observer moves)

The angle of elevation of the top of a tower is $30^{\circ}$ . After walking 40 m towards the tower it becomes $60^{\circ}$ . Find the height.

Let the height be $h$ and the final distance $x$ . Then $h = x \sqrt{3}$ and $h = \frac{x + 40}{\sqrt{3}}$ . So $3 x = x + 40$ , giving $x = 20$ and $h = 20 \sqrt{3}$ m.

Worked example 3 (elevation and depression from one point)

From the top of an 8 m building, the angle of elevation of the top of a pole is $45^{\circ}$ and the angle of depression of its foot is $30^{\circ}$ . Find the height of the pole.

Horizontal distance $d = 8 \cot 30^{\circ} = 8 \sqrt{3}$ . The part of the pole above the building's level is $d \tan 45^{\circ} = 8 \sqrt{3}$ . Pole $= 8 + 8 \sqrt{3} = 8 (1 + \sqrt{3})$ m.

Common errors

Measuring the angle of depression from the vertical instead of the horizontal.
Forgetting the observer's eye height when a person's height is given.
Using $\sin$ when the hypotenuse is not involved. Pick the ratio from the two sides you actually have.
Rounding $\sqrt{3}$ too early. Keep surds until the last line.

Board-exam tips

A correct, labelled diagram usually carries a mark in 3- and 5-mark questions. Always draw one.
Syllabus limits: at most two right triangles, and angles of elevation/depression of only 30°, 45° and 60°.
State units and give the answer in the form asked (surd or decimal).

Topics in this chapter: Angle of elevation · Angle of depression · Two-triangle problems · Angle of elevation and depression.

Formulas for this chapter: Chapter 9 on the Class 10 formula sheet (free PDF and A3 poster).

Some Applications of Trigonometry Class 10: important questions

10 original questions in the board paper's question types, from 1-mark MCQs to a case study, each with a step-marked worked answer. Try each one first. 5 worked answers open here; the other 5 open with a free account (no card).

Download these questions (PDF) 10 questions, 5 worked answers · 7 pages · 578 KB

Section A style · Multiple choice and assertion–reason (1 mark each)

Q1·1 mark·Multiple choiceAngle of elevation

From a point on level ground $25$ m from the foot of a vertical tower, the angle of elevation of its top is $60^{\circ}$ . The height of the tower is

A vertical tower of height h on level ground. From a point 25 m from its foot, the line of sight to the top makes an angle of elevation of 60°. — Tower seen from 25 m at an angle of elevation of 60°

Show worked answer

$\begin{aligned} \tan 60^{\circ} = \frac{h}{25} \\ \Rightarrow & h = 25 \sqrt{3} \end{aligned}$ m A1
Correct option: (c) $25 \sqrt{3}$ m

Q2·1 mark·Multiple choiceAngle of elevation

A pole's shadow on level ground is only $\frac{1}{\sqrt{3}}$ of the pole's own height. At that moment the sun's angle of elevation is

Q3·1 mark·Assertion–reasonAngle of elevation

In the following question, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option.

Assertion (A): As you walk in a straight line towards the foot of a tower, the angle of elevation of its top increases.

Reason (R): For angles between $0^{\circ}$ and $90^{\circ}$ , $\tan θ$ increases as $θ$ increases.

Section B style · Short answer (2 marks each)

Q4·2 marks·Short answerAngle of elevation

A straight escalator $24$ m long rises at $30^{\circ}$ to the horizontal. Through what vertical height does it carry a passenger?

A straight escalator 24 m long rising at 30° to the horizontal. The vertical height it rises, h, is shown as a dashed line at right angles to the ground. — Escalator 24 m long at 30° to the horizontal

Show worked answer

Answer:

12

$\sin 30^{\circ} = \frac{h}{24}$ M1
$h = 24 \times \frac{1}{2} = 12$ m A1

Q5·2 marks·Short answerAngle of depression

From the top of a building $45$ m high, the angle of depression of a car on the road is $60^{\circ}$ . How far is the car from the foot of the building?

A building 45 m high. From its top, a horizontal dashed line and the line of sight down to a car on the road make an angle of depression of 60°. The car's distance from the foot of the building is d. — Angle of depression of 60° from the top of a 45 m building

Section C style · Short answer (3 marks each)

Q6·3 marks·Short answerTwo-triangle problems

Two people stand on level ground on opposite sides of a vertical pole $30$ m tall, in a straight line with its foot. The angles of elevation of the top of the pole from them are $30^{\circ}$ and $60^{\circ}$ . Find the distance between them.

A vertical pole 30 m tall. Two people stand on level ground on opposite sides of its foot, in a straight line with it; the angles of elevation of the top from them are 30° (the farther one) and 60° (the nearer one). — Pole 30 m tall: angles of elevation 30° and 60°

Show worked answer

Answer:

40 \sqrt{3}

$\begin{aligned} \tan 30^{\circ} = \frac{30}{d_{1}} \\ \Rightarrow & d_{1} = 30 \sqrt{3} \end{aligned}$ m M1
$\begin{aligned} \tan 60^{\circ} = \frac{30}{d_{2}} \\ \Rightarrow & d_{2} = \frac{30}{\sqrt{3}} = 10 \sqrt{3} \end{aligned}$ m A1
Distance $= 30 \sqrt{3} + 10 \sqrt{3} = 40 \sqrt{3}$ m A1

Q7·3 marks·Short answerAngle of elevation

A drone hovers directly above a point $P$ on level ground. From a point $A$ on the ground $40$ m from $P$ , the angle of elevation of the drone is $30^{\circ}$ . The drone then rises vertically until its angle of elevation from $A$ is $60^{\circ}$ . How far does it rise?

A drone above the point P on level ground. From A, 40 m from P, its angle of elevation is first 30°; after it rises vertically the angle is 60°. The distance it rises is marked with a question mark. — From A, 40 m from P: elevation 30°, then 60°

Section D style · Long answer (5 marks each)

Q8·5 marks·Long answerAngle of depression

A straight zip-line runs from the top $A$ of a vertical tower $30$ m high down to a point $B$ on level ground, making an angle of $30^{\circ}$ with the ground. (a) Find the length of the zip-line and the distance of $B$ from the foot of the tower. (b) A rider stops at $P$ on the line, $20$ m from $A$ . How high above the ground is the rider, and how far horizontally from the tower?

A vertical tower 30 m high with a straight zip-line from its top A down to B on level ground, at 30° to the ground. A rider stops at P on the line, 20 m from A; a dashed line drops from P to the ground. — Zip-line from the top of a 30 m tower at 30° to the ground

Show worked answer

Answer:

60

30 \sqrt{3}

20

m high,

10 \sqrt{3}

m from the tower

(a) $\begin{aligned} \sin 30^{\circ} = \frac{30}{A B} \\ \Rightarrow & A B = 60 \end{aligned}$ m M1
Distance of $B$ from the foot $= 30 \sqrt{3}$ m (from $\tan 30^{\circ} = \frac{30}{d}$ ) A1
(b) Along the line the rider has dropped $20 \sin 30^{\circ} = 10$ m M1
Height above the ground $= 30 - 10 = 20$ m A1
Horizontal distance from the tower $= 20 \cos 30^{\circ} = 10 \sqrt{3}$ m A1

Q9·5 marks·Long answerTwo-triangle problems

A helicopter hovers at a point $H$ vertically above a point $O$ on level ground. From a point $A$ on the ground the angle of elevation of $H$ is $60^{\circ}$ ; from $B$ , which is $80$ m farther from $O$ along the line $O A$ , it is $30^{\circ}$ . Find the height of the helicopter and the distance $O A$ . (Use $\sqrt{3} = 1.73$ for the height in metres.)

Section E style · Case study (4 marks)

Q10·4 marks·Case studyAngle of elevation

Mobile tower. A student stands on level ground $30$ m from the foot of a vertical mobile tower and sees its top at an angle of elevation of $60^{\circ}$ .

Find the height of the tower. [1 mark]
Find the distance from the student to the top of the tower. [1 mark]
How far back, along the same line, must the student walk for the angle of elevation to become $30^{\circ}$ ? [2 marks]

A vertical mobile tower on level ground. A student 30 m from its foot sees the top at an angle of elevation of 60°. — Student 30 m from the tower, angle of elevation 60°

Show worked answer

$h = 30 \tan 60^{\circ} = 30 \sqrt{3}$ m A1
$\begin{aligned} \cos 60^{\circ} = \frac{30}{ℓ} \\ \Rightarrow & ℓ = 60 \end{aligned}$ m A1
New distance $\begin{aligned} d & = \frac{30 \sqrt{3}}{\tan 30^{\circ}} \\ = 30 \sqrt{3} \times \sqrt{3} \\ = 90 \end{aligned}$ m M1
Walk back $90 - 30 = 60$ m A1

Next: practise by Route to 95 step · the revision notes · free MCQs and case studies · all Class 10 assertion–reason questions · the Class 10 syllabus checklist.

Route to 95: four steps

Work through the steps in order. Take each checkpoint closed book, about 1.5 minutes per mark; pass at 80% to move on. Two misses in a row means going back one step.

Step 1

Secure the basics

You can draw the right triangle for a height-and-distance problem and use tan, sin or cos with 30°, 45° and 60°.

Read first: Key terms; Standard method; Worked example 1 (one triangle) 11 practice questions · checkpoint: 3 questions, 5 marks, pass 80%

Practise step 1

Step 2

Board standard

You can solve angle-of-depression problems and the standard two-triangle problems where the observer moves.

Read first: Standard method; Worked example 2 (observer moves); Worked example 3 (elevation and depression from one point) 14 practice questions · checkpoint: 3 questions, 8 marks, pass 80%

Practise step 2

Step 3

Full marks on long answers

You can write full-marks 5-mark answers with a labelled figure, two equations and a final answer with units.

Read first: Worked examples 2-3; Board-exam tips 7 practice questions · checkpoint: 3 questions, 14 marks, pass 80%

Practise step 3

Step 4

95+ stretch (HOTS)

You can handle mixed elevation-and-depression problems, moving objects and speed, and the unusual two-triangle set-ups.

Read first: Worked example 3; Common errors 5 practice questions · checkpoint: 3 questions, 9 marks, pass 80%

Practise step 4

Practice questions

Original questions in the board's styles. Multiple-choice answers are checked as you go; for written answers, compare your working with the step mark scheme and record your marks. 33 of the 45 are competency-based (case studies and questions set in a real-life situation): the "Competency-based" button shows just those.

Q1·1 mark·Multiple choiceAngle of elevation

The shadow of a vertical tower on level ground is equal to its height. The angle of elevation of the Sun is

(a) $30^{\circ}$
(b) $45^{\circ}$
(c) $60^{\circ}$
(d) $90^{\circ}$

Q2·1 mark·Multiple choiceAngle of elevation

A straight zip-line cable, 80 m long, runs from the top of a platform down to a point on level ground and makes an angle of $30^{\circ}$ with the ground. The height of the platform is

(a) $40$ m
(b) $40 \sqrt{3}$ m
(c) $80 \sqrt{3}$ m
(d) $\frac{40}{\sqrt{3}}$ m

Q3·1 mark·Multiple choiceAngle of elevation

A 14 m ladder leans against a wall and makes an angle of $60^{\circ}$ with the ground. The distance of its foot from the wall is

(a) $7 \sqrt{3}$ m
(b) $7$ m
(c) $14 \sqrt{3}$ m
(d) $\frac{7}{\sqrt{3}}$ m

Worked solution·2 marks·Very short answerAngle of elevation

A loading ramp rises from the ground to the floor of a truck, 1.2 m high, making an angle of $30^{\circ}$ with the ground. Find the length of the ramp and the horizontal distance it covers. (Take $\sqrt{3} = 1.73$ .)

Full worked solution Free, no sign-in

$\sin 30^{\circ} = \frac{1.2}{l} \Rightarrow l = 2.4$ m M1

Horizontal distance $= 1.2 \cot 30^{\circ} = 1.2 \sqrt{3} \approx 2.08$ m A1

Every question here has a full worked solution. See them all →

Stretch yourself: original geometry problems at Intermediate level (ages 13 to 16), with hints and full solutions: Problem I15 · Problem I83 · Problem I90 · Problem I97. All 28 →

Themed lesson: Eclipse geometry maths lesson (starter, activity and worked answers). All themed maths

Knowledge organiser: Some Applications of Trigonometry on one page (key facts, formulas and mistakes to avoid; printable PDF)

Where marks are lost in Some Applications of Trigonometry

No figure, or an unlabelled one. Fix: draw it every time, with heights, distances and angles marked; it usually carries B1 in 3- and 5-mark questions.
Measuring the angle of depression from the vertical. Fix: it is measured from the horizontal at the observer's eye; it equals the angle of elevation from the other end (alternate angles).
Ignoring the observer's height. Fix: subtract the eye height from the building or tower before using tan, and add it back if the question asks for the full height.
Rounding √3 too early and losing the A1. Fix: keep surds until the final line, then use the given value (1.732).
Using sin when the hypotenuse is not involved. Fix: pick the ratio from the two sides you have: opposite and adjacent means tan.
Answers without units or not in the form asked. Fix: finish with a sentence, e.g. 'Height of the tower = 20√3 m' (or '= 34.64 m' only when the question gives √3 = 1.732).

Exam strategy: how answers are step-marked · competency-based questions

Examiner Insights: common mistakes in CBSE Class 10 Maths, with fixes (our analysis of public sources) →

Some Applications of Trigonometry in our sample papers

Once step 3 is passed, test the chapter inside a full timed paper on the 2026-27 pattern (original papers by us, not official CBSE papers).

Sample paper 1 (Board standard): questions 11, 34
Sample paper 2 (Board standard): questions 13, 36
Sample paper 3 (Above board): question 34
Sample paper 4 (Above board): questions 13, 34
Sample paper 5 (95+ challenge): question 36

Want it against the clock? Take a timed 30-mark chapter test on Some Applications of Trigonometry, new questions each time, or climb the Difficulty ladder, levels 1 to 10, three questions a level (CBSE Pro or the free trial).

Some Applications of Trigonometry: common questions

What is in CBSE Class 10 Maths Chapter 9, Some Applications of Trigonometry?

The topics are: Angle of elevation, Angle of depression, Two-triangle problems, Angle of elevation and depression. This page has revision notes for each, and 45 board-style practice questions with step-marked solutions.

How many marks is Some Applications of Trigonometry worth?

It is in the Trigonometry unit, which carries 12 of the 80 theory marks. CBSE gives marks by unit, not by chapter.

Where do students lose marks in Some Applications of Trigonometry?

No figure, or an unlabelled one. Fix: draw it every time, with heights, distances and angles marked; it usually carries B1 in 3- and 5-mark questions. Measuring the angle of depression from the vertical. Fix: it is measured from the horizontal at the observer's eye; it equals the angle of elevation from the other end (alternate angles).

Are the Some Applications of Trigonometry important questions free?

Yes. The 10 important questions on this page are free to try, from 1-mark MCQs to a case study. 5 worked answers are open on the page; the others open with a free account (no card).

Are these NCERT questions?

No. Every practice question here is original, written in the board's question styles with step marks. For the textbook itself, use the NCERT link on this page.

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Revision notes

Key terms

Standard method

Worked example 1 (one triangle)

Worked example 2 (observer moves)

Worked example 3 (elevation and depression from one point)

Common errors

Board-exam tips

Some Applications of Trigonometry Class 10: important questions

Section A style · Multiple choice and assertion–reason (1 mark each)

Section B style · Short answer (2 marks each)

Section C style · Short answer (3 marks each)

Section D style · Long answer (5 marks each)

Section E style · Case study (4 marks)

Route to 95: four steps

Secure the basics

Board standard

Full marks on long answers

95+ stretch (HOTS)

Practice questions

Full worked solution Free, no sign-in

Where marks are lost in Some Applications of Trigonometry

Some Applications of Trigonometry in our sample papers

Some Applications of Trigonometry: common questions

Work through the Route to 95 with your progress saved