CBSE Class 10 · Chapter 9 · Trigonometry · 2026-27

Some Applications of Trigonometry Class 10: MCQ and case study questions

15 multiple-choice questions and 2 case-based questions on the current (rationalised) syllabus, in the style of the board paper's Sections A and E. Try each one first; the answer, a one-line reason and a worked solution open on a tap.

15 MCQs (1 mark each)
2 case studies (4 marks each)
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Multiple-choice questions

Tap an option to check it; the answer and a worked solution open. Section A of the board paper has 18 MCQs of 1 mark each, spread over all chapters.

Q1

·1 mark·Multiple choiceAngle of elevation

A tower $20 \sqrt{3}$ m tall stands on level ground. From a point $20$ m from its foot, the angle of elevation of its top is

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Answer: (c)

60^{\circ}

Why: tan θ = 20√3/20 = √3.

$\tan θ = \frac{20 \sqrt{3}}{20} = \sqrt{3} \Rightarrow θ = 60^{\circ}$ .

Q2

·1 mark·Multiple choiceAngle of elevation

A kite is flying on a taut, straight string $50$ m long that makes an angle of $30^{\circ}$ with the level ground. The height of the kite is

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Answer: (c)

25

Why: Height = string × sin 30°.

$h = 50 \sin 30^{\circ} = 50 \times \frac{1}{2} = 25$ m.

Q3

·1 mark·Multiple choiceAngle of depression

From the top of a $30$ m tall building, the angle of depression of a point on the ground is $30^{\circ}$ . The distance of the point from the foot of the building is

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Answer: (b)

30 \sqrt{3}

Why: tan 30° = 30/d, so d = 30√3.

The angle of elevation of the top from the point is also $30^{\circ}$ : $\tan 30^{\circ} = \frac{30}{d} \Rightarrow d = 30 \sqrt{3}$ m.

Q4

·1 mark·Multiple choiceAngle of elevation

A ladder leaning against a vertical wall reaches a height of $6 \sqrt{3}$ m and makes an angle of $60^{\circ}$ with the ground. The length of the ladder is

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Answer: (d)

12

Why: sin 60° = height/ladder.

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

Q5

·1 mark·Multiple choiceAngle of elevation

As the Sun's elevation rises from $30^{\circ}$ to $60^{\circ}$ , the shadow of a $12$ m tall tower becomes shorter by

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Answer: (a)

8 \sqrt{3}

Why: Shadows 12√3 and 12/√3 = 4√3; difference 8√3.

Shadow at $30^{\circ}$ : $\frac{12}{\tan 30^{\circ}} = 12 \sqrt{3}$ ; at $60^{\circ}$ : $\frac{12}{\sqrt{3}} = 4 \sqrt{3}$ . Shorter by $8 \sqrt{3}$ m.

Q6

·1 mark·Multiple choiceAngle of depression

From the top of a lighthouse $40$ m high, the angle of depression of a boat is $60^{\circ}$ . The distance of the boat from the foot of the lighthouse is

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Answer: (b)

\frac{40 \sqrt{3}}{3}

Why: tan 60° = 40/d gives d = 40/√3.

$\tan 60^{\circ} = \frac{40}{d} \Rightarrow d = \frac{40}{\sqrt{3}} = \frac{40 \sqrt{3}}{3}$ m.

Q7

·1 mark·Multiple choiceAngle of elevation

A girl whose eyes are $1.2$ m above the ground stands $18 \sqrt{3}$ m from a vertical pole. The angle of elevation of the top of the pole from her eyes is $30^{\circ}$ . The height of the pole is

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Answer: (b)

19.2

Why: The part above eye level is 18√3 tan 30° = 18 m; add the 1.2 m eye height.

Height above eye level $= 18 \sqrt{3} \tan 30^{\circ} = 18 \sqrt{3} \times \frac{1}{\sqrt{3}} = 18$ m. Pole $= 18 + 1.2 = 19.2$ m.

Q8

·1 mark·Multiple choiceProblems with two right triangles

From a point on level ground, the angle of elevation of the top of a tower is $30^{\circ}$ . After walking $40$ m towards the tower, it becomes $60^{\circ}$ . The height of the tower is

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Answer: (a)

20 \sqrt{3}

Why: h = x√3 and h = (x + 40)/√3 give x = 20.

Let the nearer point be $x$ m from the foot: $h = x \sqrt{3}$ and $h = \frac{x + 40}{\sqrt{3}}$ . So $3 x = x + 40 \Rightarrow x = 20$ , $h = 20 \sqrt{3}$ m.

Q9

·1 mark·Multiple choiceProblems with two right triangles

From the top of a $60$ m tall building, the angles of depression of the top and the foot of a lamp post are $30^{\circ}$ and $60^{\circ}$ . The height of the lamp post is

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Answer: (c)

40

Why: Distance d = 60/√3 = 20√3; the drop to the post's top is d tan 30° = 20.

$d = \frac{60}{\tan 60^{\circ}} = 20 \sqrt{3}$ m. Drop from the roof to the post's top $= d \tan 30^{\circ} = 20$ m. Post $= 60 - 20 = 40$ m.

Q10

·1 mark·Multiple choiceAngle of elevation

The angle of elevation of the top of a tree from a point $12$ m from its foot is $60^{\circ}$ . The height of the tree is

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Answer: (c)

12 \sqrt{3}

Why: h = 12 tan 60°.

$h = 12 \tan 60^{\circ} = 12 \sqrt{3}$ m.

Q11

·1 mark·Multiple choiceAngle of depression

The angle of depression of an object on the ground, seen from a point above it, is

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Answer: (a) equal to the angle of elevation of that point seen from the object

Why: The two angles are alternate angles between parallel horizontal lines.

The horizontal through the observer and the horizontal through the object are parallel, and the line of sight is a transversal, so the two angles are alternate angles and are equal.

Q12

·1 mark·Multiple choiceAngle of elevation

A straight slide in a park is $8$ m long and makes an angle of $30^{\circ}$ with the ground. The height of its top above the ground is

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Answer: (b)

4

Why: Height = 8 sin 30°.

$h = 8 \sin 30^{\circ} = 4$ m.

Q13

·1 mark·Multiple choiceAngle of elevation

A taut wire runs from the top of a vertical pole $12$ m tall to the ground and makes an angle of $30^{\circ}$ with the ground. The length of the wire is

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Answer: (d)

24

Why: sin 30° = 12/l.

$\sin 30^{\circ} = \frac{12}{l} \Rightarrow l = 24$ m.

Q14

·1 mark·Multiple choiceProblems with two right triangles

The angle of elevation of the top of a $10$ m tall tower from a point on the ground is $45^{\circ}$ . If the point moves a further $10 (\sqrt{3} - 1)$ m away from the tower, the angle of elevation becomes

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Answer: (d)

30^{\circ}

Why: The new distance is 10 + 10(√3 − 1) = 10√3, so tan θ = 1/√3.

First distance $= 10$ m ( $\tan 45^{\circ} = 1$ ). New distance $= 10 \sqrt{3}$ m, so $\tan θ = \frac{10}{10 \sqrt{3}} = \frac{1}{\sqrt{3}}$ , $θ = 30^{\circ}$ .

Q15

·1 mark·Multiple choiceProblems with two right triangles

From the top of a $50$ m high cliff, the angles of depression of two boats in line with the foot of the cliff, on the same side, are $45^{\circ}$ and $30^{\circ}$ . The distance between the boats is

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Answer: (b)

50 (\sqrt{3} - 1)

Why: Distances 50 and 50√3 from the foot; subtract.

Nearer boat: $\frac{50}{\tan 45^{\circ}} = 50$ m; farther boat: $\frac{50}{\tan 30^{\circ}} = 50 \sqrt{3}$ m. Gap $= 50 (\sqrt{3} - 1)$ m.

Case study questions

Section E of the board paper has three case-based questions of 4 marks: a real-life passage, then parts of 1, 1 and 2 marks, with a choice (OR) on the 2-mark part. All Class 10 case study questions by chapter.

Case study 1: Lighthouse watch (4 marks)

A coastguard stands at the top of a lighthouse $45$ m above sea level. She watches a fishing boat sailing straight towards the foot of the lighthouse. At first the angle of depression of the boat is $30^{\circ}$ ; some time later it is $60^{\circ}$ .

(i) How far is the boat from the foot of the lighthouse when the angle of depression is

30^{\circ}

? [1 mark]

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Answer:

45 \sqrt{3}

\tan 30^{\circ} = \frac{45}{d_{1}} \Rightarrow d_{1} = 45 \sqrt{3}

m. A1

(ii) How far is it when the angle of depression is

60^{\circ}

? [1 mark]

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Answer:

15 \sqrt{3}

d_{2} = \frac{45}{\tan 60^{\circ}} = \frac{45}{\sqrt{3}} = 15 \sqrt{3}

m. A1

(iii) Find the distance the boat travels between the two observations. [2 marks]

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Answer:

30 \sqrt{3}

d_{1} - d_{2} = 45 \sqrt{3} - 15 \sqrt{3}

= 30 \sqrt{3}

m. A1

OR The boat takes

3

minutes between the two observations. Find its speed in metres per minute. (Use

\sqrt{3} = 1.73

.) [2 marks]

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Answer:

17.3

m/min

Distance

= 30 \sqrt{3} = 30 \times 1.73 = 51.9

m M1; speed

= 51.9 \div 3 = 17.3

m/min. A1

Case study 2: Drone at the sports day (4 marks)

At a school sports day, a drone films the events while hovering vertically above a point $P$ of the level playground. Ishaan stands at a point $A$ , $40$ m from $P$ . At first the angle of elevation of the drone from $A$ is $45^{\circ}$ . The drone then rises vertically and the angle of elevation from $A$ becomes $60^{\circ}$ . (Ignore Ishaan's height.)

(i) Find the first height of the drone. [1 mark]

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Answer:

40

h_{1} = 40 \tan 45^{\circ} = 40

m. A1

(ii) Find the height of the drone after it rises. [1 mark]

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Answer:

40 \sqrt{3}

h_{2} = 40 \tan 60^{\circ} = 40 \sqrt{3}

m. A1

(iii) How far did the drone rise? [2 marks]

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Answer:

40 (\sqrt{3} - 1)

h_{2} - h_{1} = 40 \sqrt{3} - 40

= 40 (\sqrt{3} - 1)

m. A1

OR Find the straight-line distance from

A

to the drone in its higher position. [2 marks]

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Answer:

80

\cos 60^{\circ} = \frac{40}{A D}

\Rightarrow A D = \frac{40}{1 / 2} = 80

m. A1

Next steps for Some Applications of Trigonometry

This free set is separate from the chapter's question bank. On the Some Applications of Trigonometry chapter page the revision notes and the first step of the Route to 95 are free for everyone. CBSE Pro adds all 45 questions in the chapter bank (short and long answers, assertion–reason and more case studies) with their full step-marking schemes, the Route to 95 checkpoints with your progress saved, Skill Builders, the full common-mistakes library and AI marking of your written working.

Some Applications of Trigonometry: important questions What Pro includes

Original questions written by CBSE Math Revision for the CBSE 2026-27 syllabus; not taken from NCERT, NCERT Exemplar or CBSE papers. Every answer was checked twice, the second time by an independent reviewer. CBSE Math Revision is independent and not affiliated with CBSE or NCERT. Spotted a slip? Tell us and it goes in the corrections log.