A drone was used to facilitate movement of an ambulance on the straight highway to a point P on the ground where there was an accident. The ambulance was travelling at the speed of 60 km/h. The drone stopped at a point Q, 100 m vertically above the point P. The angle of depression of the ambulance was found to be 30° at a particular instant. Based on above information, answer the following questions : (i) Represent the above situation with the help of a diagram. (ii) Find the distance between the ambulance and the site of accident (P) at the particular instant. (Use √(3) = 1.73) (iii)(a) Find the time (in seconds) in which the angle of depression changes from 30° to 45°. (iii)(b) How long (in seconds) will the ambulance take to reach point P from a point T on the highway such that angle of depression of the ambulance at T is 60° from the drone ?

MathsCBSE Class 10 · Standard · Board exam 2025, set 30/4/1 · 4 marks · Case-based

CBSE Class 10 Mathematics Standard 2025: question 38

Set in the CBSE Class 10 board exam of 2025, paper 30/4/1, section E, question 38, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q38 · 4 marks · Case-basedCompetency-basedSome Applications of Trigonometry

A drone was used to facilitate movement of an ambulance on the straight highway to a point P on the ground where there was an accident. The ambulance was travelling at the speed of 60 km/h. The drone stopped at a point Q, 100 m vertically above the point P. The angle of depression of the ambulance was found to be $3 0^{\circ}$ at a particular instant.

Based on above information, answer the following questions :

(i)
Represent the above situation with the help of a diagram.
1 mark
(ii)
Find the distance between the ambulance and the site of accident (P) at the particular instant. (Use $3 = 1.73$ )
1 mark
(iii)(a)
Find the time (in seconds) in which the angle of depression changes from $3 0^{\circ}$ to $4 5^{\circ}$ .
2 marks
OR
(iii)(b)
How long (in seconds) will the ambulance take to reach point P from a point T on the highway such that angle of depression of the ambulance at T is $6 0^{\circ}$ from the drone ?
2 marks

Figure in the paper: Drone at point Q, vertically 100 m above accident point P on the ground. An ambulance on the straight highway is shown to the right, with a line of sight from the drone to the ambulance making the angle of depression at Q. A police car is also shown on the highway between P and the ambulance. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 30/4/1.

Hints

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

Hint 1

The angle of depression at Q equals the angle of elevation at the ambulance [alternate angles], giving a right triangle with height 100 m.

Hint 2

Use tan (angle) = height / horizontal distance for each position of the ambulance.

Hint 3

Convert 60 km/h into m/s by multiplying by $\frac{5}{18}$ , then use time = distance ÷ speed.

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.

(i) Diagram: draw a vertical segment QP of length 100 m, with Q the drone and P the accident site on the ground. Draw a horizontal ground line from P with the ambulance at R on it, so that $∠ QP R = 9 0^{\circ}$ and $P R = d$ . Draw a horizontal ray QS through Q and join QR. Mark $∠ S QR = 3 0^{\circ}$ (angle of depression), which equals $∠ QR P = 3 0^{\circ}$ [alternate angles].
1 mark
(ii) In right $Δ QP R$ : $tan 3 0^{\circ} = \frac{QP}{P R} \Rightarrow \frac{100}{d} = \frac{1}{3}$
½ mark
(ii) $d = 1003 = 100 \times 1.73 = 173$ m
½ mark
(iii)(a) Figure: mark a point M on PR where the angle of depression is $4 5^{\circ}$ , so $∠ QM P = 4 5^{\circ}$ . Let $R M = x$ , so that $P M = 173 - x$ .
½ mark
(iii)(a) In right $Δ QP M$ : $tan 4 5^{\circ} = \frac{100}{173 - x} = 1$
½ mark
(iii)(a) $173 - x = 100 \Rightarrow x = 73$ m
½ mark
(iii)(a) Speed $= 60 \times \frac{5}{18} = \frac{50}{3}$ m/s, so time $= \frac{73}{50/3} = \frac{219}{50} \approx 4.4$ s.
½ mark
(iii)(b) Figure: mark a point T on the highway with $∠ QT P = 6 0^{\circ}$ (angle of depression $6 0^{\circ}$ ), and let $P T = y$ .
½ mark
(iii)(b) In right $Δ QP T$ : $tan 6 0^{\circ} = \frac{100}{y} = 3$
½ mark
(iii)(b) $y = \frac{100}{3} = \frac{100 3}{3} = \frac{173}{3}$ m $\approx 57.67$ m
½ mark
(iii)(b) Time $= \frac{173/3}{50/3} = \frac{173}{50} \approx 3.5$ s (exactly $23$ s).
½ mark

Answer: (ii) 173 m; (iii)(a) ≈ 4.4 s (219/50 s); (iii)(b) 2√3 ≈ 3.5 s

Where the marks are

(i) Correct labelled figure showing QP = 100 m, a right angle at P and the $3 0^{\circ}$ angle of depression (1 mark)
(ii) Uses $tan 3 0^{\circ} = \frac{100}{d}$ (½ mark)
(ii) Gets $d = 173$ m (½ mark)
(iii)(a) Correct figure with the $4 5^{\circ}$ position marked (½ mark)
(iii)(a) Forms $\frac{100}{173 - x} = 1$ (½ mark)
(iii)(a) Gets $x = 73$ m (½ mark)
(iii)(a) Converts the speed and finds time ≈ 4.4 s (½ mark)
(iii)(b) Correct figure with the $6 0^{\circ}$ position (½ mark)
(iii)(b) Forms $\frac{100}{y} = 3$ (½ mark)
(iii)(b) Gets $y = \frac{173}{3}$ m (½ mark)
(iii)(b) Time $= 23 \approx 3.5$ s (½ mark)

Revise the chapter

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

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 paper2025 board paper, set 30/4/1 (Standard): every question, by section.

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CBSE Class 10 Mathematics Standard 2025: question 38

Hints

Step-by-step solution

Revise the chapter

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