A kite is flying at a height of 60 m above the ground level. Ravi, standing at the roof of the house is holding the string straight and observes the angle of elevation of kite as 30°. From the bottom of the same building, the angle of elevation of kite is 45°. Find the length of the string and height of roof from the ground. (Use √(3) = 1.73)

MathsCBSE Class 10 · Standard · Board exam 2026, set 30/4/1 · 5 marks · Long answer

CBSE Class 10 Mathematics Standard 2026: question 34

Set in the CBSE Class 10 board exam of 2026, paper 30/4/1, section D, question 34, for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q34 · 5 marks · Long answerCompetency-basedSome Applications of Trigonometry

A kite is flying at a height of 60 m above the ground level. Ravi, standing at the roof of the house is holding the string straight and observes the angle of elevation of kite as $3 0^{\circ}$ . From the bottom of the same building, the angle of elevation of kite is $4 5^{\circ}$ . Find the length of the string and height of roof 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 30/4/1.

Hints

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

Hint 1

Draw the kite's height as a vertical line and the building as a shorter vertical line; join the roof to the kite and to the kite's vertical.

Hint 2

The 45° angle from the bottom gives the horizontal distance between the building and the kite.

Hint 3

Use $tan 3 0^{\circ}$ from the roof to find the part of the kite's height above roof level, then $sin 3 0^{\circ}$ for the string.

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: let K be the kite and G the point on the ground directly below it, so KG = 60 m. TR is the building, with T its bottom and R the roof where Ravi stands. Draw RS ⟂ KG with S on KG, so SG = TR and SR = GT. The angle of elevation of K from R is $∠ K R S = 3 0^{\circ}$ , and from T it is $∠ K T G = 4 5^{\circ}$ .
1 mark
In right $△ K GT$ : $tan 4 5^{\circ} = \frac{K G}{GT} \Rightarrow 1 = \frac{60}{GT} \Rightarrow GT = 60$ m, so $S R = 60$ m.
1 mark
In right $△ K S R$ : $tan 3 0^{\circ} = \frac{K S}{S R} \Rightarrow \frac{1}{3} = \frac{K S}{60}$
1 mark
$K S = \frac{60}{3} = 203 = 20 \times 1.73 = 34.6$ m
½ mark
Height of roof $T R = S G = K G - K S = 60 - 34.6 = 25.4$ m.
½ mark
Length of string KR: $sin 3 0^{\circ} = \frac{K S}{K R} \Rightarrow \frac{1}{2} = \frac{20 3}{K R}$
½ mark
$K R = 403 = 40 \times 1.73 = 69.2$ m. So the string is 69.2 m long and the roof is 25.4 m above the ground.
½ mark

Answer: Length of string = 69.2 m; height of roof = 25.4 m

Where the marks are

Correct labelled figure (1 mark)
Uses $tan 4 5^{\circ}$ to get GT = 60 m (1 mark)
Uses $tan 3 0^{\circ} = \frac{K S}{S R}$ (1 mark)
KS = $203$ = 34.6 m (½ mark)
Height of roof = 25.4 m (½ mark)
Uses $sin 3 0^{\circ} = \frac{K S}{K R}$ (½ mark)
String length = 69.2 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 paper2026 board paper, set 30/4/1 (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 34

Hints

Step-by-step solution

Revise the chapter

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