MathsCBSE Class 10 · Standard · Sample paper, session 2026-27 · 5 marks · Long answer

CBSE Class 10 Mathematics Standard sample paper 2026-27: question 35(A)

From the official CBSE sample question paper for 2026-27, section D, question 35(A), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q35(A) · 5 marks · Long answerCompetency-basedSome Applications of Trigonometry

Manish is standing on level ground and observes a kite flying 200 m away from him at an angle of elevation of $3 0^{\circ}$ . Mahesh, standing on the roof of a 50 m high building on the opposite side of the kite, observes the same kite at an angle of elevation of $4 5^{\circ}$ . Find the distance between the kite and Mahesh.

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2026-27.

Hints

Open one hint at a time and try again before the next.

Hint 1

200 m is the line-of-sight distance (hypotenuse), so use sine to find the kite's height.

Hint 2

Subtract the 50 m building height from the kite's height to get the vertical rise seen from Mahesh.

Hint 3

Use $sin 4 5^{\circ}$ in the small right triangle at Mahesh.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

Figure (in words):
- Manish stands at A on the ground.
- The kite is at B, with BE ⟂ ground and E on the ground.
- The building is DC = 50 m, with Mahesh at D on its roof, on the other side of E.
- R is the point on BE with DR ∥ CE, so RE = DC = 50 m.
- AB = 200 m and ∠BAE = 30°.
- ∠BDR = 45°.
1 mark
In right $△ A E B$ : $\frac{B E}{A B} = sin 3 0^{\circ} = \frac{1}{2}$
1 mark
$\Rightarrow B E = \frac{200}{2} = 100$ m
½ mark
$B R = B E - R E = B E - D C = 100 - 50 = 50$ m
1 mark
In right $△ B R D$ : $\frac{B R}{B D} = sin 4 5^{\circ} \Rightarrow \frac{50}{B D} = \frac{1}{2}$
1 mark
$\Rightarrow B D = 502$ m $\approx 70.7$ m
So the distance between the kite and Mahesh is $502$ m.
½ mark

Answer: $502$ m (≈ 70.7 m)

Where the marks are

Draws a correct labelled figure (1 mark)
Uses sin 30° in triangle AEB (1 mark)
Gets BE = 100 m (½ mark)
Finds BR = 50 m by subtracting the building height (1 mark)
Uses sin 45° in triangle BRD (1 mark)
Gets BD = 50√2 m (½ mark)

Revise the chapter

Chapter: Some Applications of Trigonometry (notes, formulas, where marks are lost and every past question)
Topic: Line of Sight and Angle of Elevation

Practise this chapterBoard-style questions from Some Applications of Trigonometry, one at a time, marked against the answer key and marking points.Back to the full paperSample paper 2026-27 (Standard): every question, by section.

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.

CBSE Class 10 Mathematics Standard sample paper 2026-27: question 35(A)

Hints

Step-by-step solution

Revise the chapter

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