MathsCBSE Class 10 · Standard · Sample paper, session 2024-25 · 5 marks · Long answer

CBSE Class 10 Mathematics Standard sample paper 2024-25: question 34

From the official CBSE sample question paper for 2024-25, 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 boy whose eye level is $1.35$ m from the ground, spots a balloon moving with the wind in a horizontal line at some height from the ground. The angle of elevation of the balloon from the eyes of the boy at an instant is $6 0^{\circ}$ . After $12$ seconds, the angle of elevation reduces to $3 0^{\circ}$ . If the speed of the wind is $3$ m/s then find the height of the balloon from the ground. (Use $3 = 1.73$ )

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2024-25.

Hints

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

Hint 1

Measure every height from the boy's eye level first, then add 1.35 m at the end.

Hint 2

The balloon moves horizontally $3 \times 12$ m between the two sightings, so the second horizontal distance is $x + 36$ .

Hint 3

Write $tan 6 0^{\circ}$ and $tan 3 0^{\circ}$ in terms of $h$ and $x$ , then eliminate $h$ .

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 $A$ be the boy's eye, with $A D = 1.35$ m the eye height above the ground. $B$ and $C$ are the two positions of the balloon, both on a horizontal line at height $h$ above eye level. $G$ and $F$ are the feet of the perpendiculars from $B$ and $C$ to the horizontal line through $A$ . Then $∠ B A G = 6 0^{\circ}$ , $∠ C A F = 3 0^{\circ}$ , $A G = x$ and $B G = C F = h$ .
1 mark
Distance moved by the balloon in 12 s $= 3 \times 12 = 36$ m.
So $B C = GF = 36$ m and $A F = x + 36$ .
1 mark
In right $△ A GB$ : $tan 6 0^{\circ} = \frac{h}{x}$ , so $3 = \frac{h}{x}$ and $h = 3 x$ ...(i)
1 mark
In right $△ A F C$ : $tan 3 0^{\circ} = \frac{h}{x + 36}$ , so $\frac{1}{3} = \frac{h}{x + 36}$ and $h = \frac{x + 36}{3}$ ...(ii)
1 mark
From (i) and (ii): $3 x = \frac{x + 36}{3}$ .
This gives $3 x = x + 36$ , so $x = 18$ m.
Then $h = 183 = 18 \times 1.73 = 31.14$ m.
Height of the balloon from the ground $= 31.14 + 1.35 = 32.49$ m.
1 mark

Answer: Height of balloon from ground = 18√3 + 1.35 ≈ 32.49 m

Where the marks are

Draws a correct figure with the eye level, both balloon positions and both angles (1 mark)
Finds the distance moved $= 36$ m and sets $B C = GF = 36$ m (1 mark)
Forms $h = 3 x$ from $tan 6 0^{\circ}$ (1 mark)
Forms $h = (x + 36) / 3$ from $tan 3 0^{\circ}$ (1 mark)
Solves to get $h = 183 = 31.14$ m and adds 1.35 m to get 32.49 m (1 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 2024-25 (Standard): every question, by section.

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CBSE Class 10 Mathematics Standard sample paper 2024-25: question 34

Hints

Step-by-step solution

Revise the chapter

More questions from Some Applications of Trigonometry