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

CBSE Class 10 Mathematics Standard 2025: question 38

Set in the CBSE Class 10 board exam of 2025, paper 30/1/1, 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

Case Study – 3
Amrita stood near the base of a lighthouse, gazing up at its towering height. She measured the angle of elevation to the top and found it to be $6 0^{\circ}$ . Then, she climbed a nearby observation deck, 40 metres higher than her original position and noticed the angle of elevation to the top of lighthouse to be $4 5^{\circ}$ .

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

(i)
If CD is h metres, find the distance BD in terms of 'h'.
1 mark
(ii)
Find distance BC in terms of 'h'.
1 mark
(iii)(a)
Find the height CE of the lighthouse [Use $3 = 1 \cdot 73$ ]
2 marks
OR
(iii)(b)
Find distance AE, if AC = 100 m.
2 marks

Figure in the paper: Photo of a lighthouse on a cliff, and a diagram: vertical lighthouse CE with C at top, E at base; D lies on CE with CD = h. A is on the ground to the left of E, B is directly above A with AB = 40 m, and BD is horizontal to D. Angle of elevation at B to C is 45° and at A to C (line AC) is 60°. See it in the official paper.

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

Hints

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

Hint 1

Look at the right triangle BDC with the 45° angle at B.

Hint 2

ABDE is a rectangle, so AE = BD and DE = 40 m.

Hint 3

In triangle AEC, use tan 60° for (iii)(a) and cos 60° for (iii)(b).

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) In right $△ B D C$ (right angle at D): $tan 4 5^{\circ} = \frac{C D}{B D} = \frac{h}{B D}$
½ mark
(i) Since $tan 4 5^{\circ} = 1$ , $B D = h$ m.
½ mark
(ii) In $△ B D C$ : $sin 4 5^{\circ} = \frac{C D}{B C}$ , so $\frac{1}{2} = \frac{h}{B C}$
½ mark
(ii) $B C = 2 h$ m
½ mark
(iii)(a) ABDE is a rectangle, so $A E = B D = h$ and $D E = A B = 40$ m. This gives $C E = h + 40$ .
In right $△ A E C$ : $tan 6 0^{\circ} = \frac{C E}{A E}$ , so $3 = \frac{h + 40}{h}$
1 mark
(iii)(a) $3 h - h = 40$ , so $h = \frac{40}{3 - 1} = \frac{40 ( 3 + 1 )}{2} = 20 (3 + 1) = 20 \times 2.73 = 54.6$ m
½ mark
(iii)(a) $C E = 54.6 + 40 = 94.6$ m
½ mark
(iii)(b) In right $△ A E C$ : $cos 6 0^{\circ} = \frac{A E}{A C}$ , so $\frac{1}{2} = \frac{A E}{100}$
1 mark
(iii)(b) $A E = 50$ m
1 mark

Answer: (i) BD = h m; (ii) BC = √2 h m; (iii)(a) CE = 94.6 m, (iii)(b) AE = 50 m

Where the marks are

(i) Uses $tan 4 5^{\circ} = h / B D$ (½ mark)
(i) Gets $B D = h$ (½ mark)
(ii) Uses $sin 4 5^{\circ} = h / B C$ (½ mark)
(ii) Gets $B C = 2 h$ (½ mark)
(iii)(a) Forms $3 = \frac{h + 40}{h}$ (1 mark)
(iii)(a) Finds $h = 20 (3 + 1) = 54.6$ m (½ mark)
(iii)(a) Gets $C E = 94.6$ m (½ mark)
(iii)(b) Uses $cos 6 0^{\circ} = A E / A C$ (1 mark)
(iii)(b) Gets $A E = 50$ m (1 mark)

Revise the chapter

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

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 30/1/1 (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 38

Hints

Step-by-step solution

Revise the chapter

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