A boy standing on a horizontal plane is flying a kite with a string of length 60 m, at an angle of elevation of 30°. Another boy standing on the roof of a 20 m high building, finds the angle of elevation of same kite to be 45°. If both the boys are on opposite sides of the kite, find the distance of the first boy from the base of the building. Also, find the height of the kite from the ground. (Use √(3) = 1.73)

MathsCBSE Class 10 · Standard · Board exam 2026, set 30/3/1 · 5 marks · Long answer

CBSE Class 10 Mathematics Standard 2026: question 35

Set in the CBSE Class 10 board exam of 2026, paper 30/3/1, section D, question 35, for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q35 · 5 marks · Long answerCompetency-basedSome Applications of Trigonometry

A boy standing on a horizontal plane is flying a kite with a string of length 60 m, at an angle of elevation of $3 0^{\circ}$ . Another boy standing on the roof of a 20 m high building, finds the angle of elevation of same kite to be $4 5^{\circ}$ . If both the boys are on opposite sides of the kite, find the distance of the first boy from the base of the building. Also, find the height of the kite from the ground. (Use $3 = 1.73$ )

Source: official question paper from cbse.gov.in (PDF). See it in the full 2026 board paper, set 30/3/1.

Hints

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

Hint 1

The string is the hypotenuse of a right triangle with angle 30° at the first boy.

Hint 2

Use sin 30° to get the kite's height and cos 30° to get the first boy's horizontal distance from the foot of the kite.

Hint 3

From the roof, only (kite height − 20) is the vertical side; tan 45° = 1 then gives the horizontal gap from the building.

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:
- K is the kite and B is the first boy on the ground, with string $K B = 60$ m and $∠ K B P = 3 0^{\circ}$ .
- CM is the 20 m building, with the second boy at M.
- KP $⊥$ ground at P, and $M L ∥ C P$ meets KP at L, so $L P = M C = 20$ m and $M L = C P$ .
- Let $K L = h$ , $C P = M L = x$ and $P B = y$ . The angle of elevation $∠ K M L = 4 5^{\circ}$ .
- C (building) and B (first boy) lie on opposite sides of P.
1 mark
In right $Δ K P B$ :
$\frac{K P}{K B} = sin 3 0^{\circ} \Rightarrow \frac{h + 20}{60} = \frac{1}{2}$
½ mark
$\Rightarrow h + 20 = 30 \Rightarrow h = 10$ m
½ mark
Also, in $Δ K P B$ :
$\frac{P B}{K B} = cos 3 0^{\circ} \Rightarrow \frac{y}{60} = \frac{3}{2}$
½ mark
$\Rightarrow y = 303 = 30 \times 1.73 = 51.9$ m
½ mark
In right $Δ K L M$ :
$\frac{K L}{M L} = tan 4 5^{\circ} \Rightarrow \frac{10}{x} = 1$
½ mark
$\Rightarrow x = 10$ m
½ mark
Height of the kite from the ground: $K P = h + 20 = 10 + 20 = 30$ m.
½ mark
Distance of the first boy from the base of the building: $C B = C P + P B = 10 + 51.9 = 61.9$ m.
½ mark

Answer: Distance of first boy from base of building = 61.9 m; height of kite = 30 m

Where the marks are

Correct labelled figure (1 mark)
Uses $sin 3 0^{\circ} = \frac{h + 20}{60}$ (½ mark) and gets $h = 10$ (½ mark)
Uses $cos 3 0^{\circ} = \frac{y}{60}$ (½ mark) and gets $y = 51.9$ m (½ mark)
Uses $tan 4 5^{\circ} = \frac{10}{x}$ (½ mark) and gets $x = 10$ (½ mark)
Height of kite 30 m (½ mark)
Distance 61.9 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 paper2026 board paper, set 30/3/1 (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 2026: question 35

Hints

Step-by-step solution

Revise the chapter

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