Case Study 2. A lighthouse stands tall on a cliff by the sea, watching over ships that pass by. One day a ship is seen approaching the shore and from the top of the lighthouse, the angles of depression of the ship are observed to be 30° and 45° as it moves from point P to point Q. The height of the lighthouse is 50 metres. Based on the information given above, answer the following questions : (i) Find the distance of the ship from the base of the lighthouse when it is at point Q, where the angle of depression is 45°. (ii) Find the measures of ∠ PBA and ∠ QBA. (iii) (a) Find the distance travelled by the ship between points P and Q. (iii) (b) If the ship continues moving towards the shore and takes 10 minutes to travel from Q to A, calculate the speed of the ship in km/h, from Q to A.

MathsCBSE Class 10 · Standard · Board exam 2025, set 30/3/1 · 4 marks · Case-based

CBSE Class 10 Mathematics Standard 2025: question 37

Set in the CBSE Class 10 board exam of 2025, paper 30/3/1, 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. A lighthouse stands tall on a cliff by the sea, watching over ships that pass by. One day a ship is seen approaching the shore and from the top of the lighthouse, the angles of depression of the ship are observed to be $3 0^{\circ}$ and $4 5^{\circ}$ as it moves from point P to point Q. The height of the lighthouse is 50 metres.

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

(i)
Find the distance of the ship from the base of the lighthouse when it is at point Q, where the angle of depression is $4 5^{\circ}$ .
1 mark
(ii)
Find the measures of $∠$ PBA and $∠$ QBA.
1 mark
(iii)
(a) Find the distance travelled by the ship between points P and Q.
2 marks
OR
(iii)
(b) If the ship continues moving towards the shore and takes 10 minutes to travel from Q to A, calculate the speed of the ship in km/h, from Q to A.
2 marks

Figure in the paper: A photograph of a lighthouse on a rocky cliff by the sea, and a line diagram: B is the top of the lighthouse and A is its base, with BA vertical on the right. P and Q lie on the horizontal ground to the left of A, with Q nearer to A. A horizontal line BX goes from B to the left. The angle between BX and BP is $3 0^{\circ}$ and the angle between BX and BQ is $4 5^{\circ}$ (both angles of depression at B). The angle at P between the ground and PB is marked. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 30/3/1.

Hints

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

Hint 1

The angle of depression from B equals the angle of elevation at the ship (alternate angles).

Hint 2

Use tan in the right triangles with AB = 50 m to find AQ and AP. PQ is their difference.

Hint 3

For speed, change metres to km and minutes to hours before dividing.

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) $B X ∥ P A$ , so $∠ B Q A = ∠ X B Q = 4 5^{\circ}$ (alternate angles). In right $△ B A Q$ : $tan 4 5^{\circ} = \frac{A B}{A Q} = \frac{50}{A Q}$
½ mark
(i) $1 = \frac{50}{A Q} \Rightarrow A Q = 50$ m
½ mark
(ii) $∠ X B A = 9 0^{\circ}$ . So $∠ P B A = 9 0^{\circ} - 3 0^{\circ} = 6 0^{\circ}$ .
½ mark
(ii) $∠ QB A = 9 0^{\circ} - 4 5^{\circ} = 4 5^{\circ}$ .
½ mark
(iii)(a) $∠ B P A = 3 0^{\circ}$ (alternate angles). In right $△ B A P$ : $tan 3 0^{\circ} = \frac{50}{A P} \Rightarrow \frac{1}{3} = \frac{50}{A P}$
½ mark
(iii)(a) $A P = 503$ m
½ mark
(iii)(a) $P Q = A P - A Q = 503 - 50 = 50 (3 - 1)$ m $\approx 50 \times 0.732 = 36.6$ m
1 mark
(iii)(b) Distance $Q A = 50$ m $= 0.05$ km. Time $= 10$ min $= \frac{1}{6}$ h. Speed $= \frac{distance}{time} = \frac{0.05}{1/6}$
1 mark
(iii)(b) $= 0.3$ km/h
1 mark

Answer: (i) 50 m; (ii) ∠PBA = 60°, ∠QBA = 45°; (iii)(a) $50 (3 - 1)$ m ≈ 36.6 m; (iii)(b) 0.3 km/h

Where the marks are

(i) Uses $tan 4 5^{\circ} = \frac{50}{A Q}$ (½ mark)
(i) Gets AQ = 50 m (½ mark)
(ii) ∠PBA = 60° (½ mark)
(ii) ∠QBA = 45° (½ mark)
(iii)(a) Uses $tan 3 0^{\circ} = \frac{50}{A P}$ (½ mark)
(iii)(a) Gets AP = $503$ m (½ mark)
(iii)(a) Gets PQ = $50 (3 - 1)$ m (1 mark)
(iii)(b) Writes speed as 50 m ÷ 10 min with the units converted (1 mark)
(iii)(b) Gets 0.3 km/h (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 paper2025 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 2025: question 37

Hints

Step-by-step solution

Revise the chapter

More questions from Some Applications of Trigonometry