MathsCBSE Class 10 · Standard · Board exam 2026, set 30/2/1 · 4 marks · Case-based

CBSE Class 10 Mathematics Standard 2026: question 37

Set in the CBSE Class 10 board exam of 2026, paper 30/2/1, section E, question 37, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q37 · 4 marks · Case-basedCompetency-basedSome Applications of Trigonometry

Case Study – 2

Tejas is standing at the top of a building and observes a car at an angle of depression of $3 0^{\circ}$ as it approaches the base of the building at a uniform speed. 6 seconds later, the angle of depression increases to $6 0^{\circ}$ , and at that moment, the car is 25 m away from the building.

The figure shows a building on the left with a car at ground level on the far left. On the right, A is the top of the building, B is the base of the building (right angle at B), and the ground line runs through D, C and B. Angle of depression $3 0^{\circ}$ from A to the first car position on the line AD... more precisely: from A, a line goes down to the car's first position at D making $6 0^{\circ}$ with the horizontal dashed line, and a line to the second position C making $3 0^{\circ}$ with the horizontal dashed line. CB = 25 m.

Based on the information given above, answer the following questions :

(i)
What is the height of the building?
1 mark
(ii)
What is the distance between the two positions of the car?
1 mark
(iii)(a)
What would be the total time taken by the car to reach the foot of the building from the starting point?
2 marks
OR
(iii)(b)
What is the distance of the observer from the car when it makes an angle of $6 0^{\circ}$ ?
2 marks

Figure in the paper: Left: a brick building (tall rectangle) with a small car on the ground at far left. Right: a right-angled figure with observer A at top right, vertical side AB down to base B (right angle at B). Ground line D—C—B. Dashed horizontal line from A; angle of depression $3 0^{\circ}$ between the horizontal and line A to C, and $6 0^{\circ}$ between the horizontal and line A to D. CB is marked 25 m. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2026 board paper, set 30/2/1.

Hints

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

Hint 1

The angle of depression from A equals the angle of elevation at the car's position (alternate angles).

Hint 2

The nearer position C goes with $6 0^{\circ}$ and the farther position D with $3 0^{\circ}$ . Use tan in each right triangle with AB as the opposite side.

Hint 3

For the time, find the speed from 50 m in 6 s, then apply it to the full distance DB.

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.

Let AB be the building, with A the observer at the top and B its foot. D is the first position of the car and C the second, with $C B = 25$ m. Using alternate angles with the horizontal through A, $∠ A D B = 3 0^{\circ}$ and $∠ A C B = 6 0^{\circ}$ .
(i) In right $△ A B C$ : $tan 6 0^{\circ} = \frac{A B}{B C} \Rightarrow 3 = \frac{A B}{25}$ .
½ mark
(i) So $A B = 253$ m. The height of the building is $253$ m (about 43.3 m).
½ mark
(ii) In right $△ A B D$ : $tan 3 0^{\circ} = \frac{A B}{B D} \Rightarrow \frac{1}{3} = \frac{25 3}{B D}$ , so $B D = 75$ m.
½ mark
(ii) Distance between the two positions: $C D = B D - B C = 75 - 25 = 50$ m.
½ mark
(iii)(a) The car covers 50 m in 6 s at uniform speed. Time to cover the whole distance $D B = 75$ m is $\frac{6}{50} \times 75$ .
1 mark
(iii)(a) $\frac{6}{50} \times 75 = 9$ s. The car takes 9 seconds in total to reach the foot of the building.
1 mark
(iii)(b) In right $△ A B C$ : $cos 6 0^{\circ} = \frac{B C}{A C} \Rightarrow \frac{1}{2} = \frac{25}{A C}$ .
1 mark
(iii)(b) So $A C = 50$ m. When the angle of depression is $6 0^{\circ}$ , the observer is 50 m from the car.
1 mark

Answer: (i) $253$ m; (ii) 50 m; (iii)(a) 9 s; (iii)(b) 50 m

Where the marks are

(i) Uses $tan 6 0^{\circ} = A B /25$ (½ mark)
(i) Gets $A B = 253$ m (½ mark)
(ii) Uses $tan 3 0^{\circ} = 253 / B D$ to get $B D = 75$ m (½ mark)
(ii) Gets $C D = 50$ m (½ mark)
(iii)(a) Sets up time for 75 m as $\frac{6}{50} \times 75$ using uniform speed (1 mark)
(iii)(a) Gets 9 s (1 mark)
(iii)(b) Uses $cos 6 0^{\circ} = B C / A C$ (1 mark)
(iii)(b) Gets $A C = 50$ m (1 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 paper2026 board paper, set 30/2/1 (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 37

Hints

Step-by-step solution

Revise the chapter

More questions from Some Applications of Trigonometry