As observed from the top of a 75 m high lighthouse from the sea level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships (Use √(3) = 1.732)

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

CBSE Class 10 Mathematics Basic sample paper 2024-25: question 35(B)

From the official CBSE sample question paper for 2024-25, section D, question 35(B), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q35(B) · 5 marks · Long answerSome Applications of Trigonometry

As observed from the top of a 75 m high lighthouse from the sea level, the angles of depression of two ships are $3 0^{\circ}$ and $4 5^{\circ}$ . If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships (Use $3 = 1.732$ )

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

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

Hint 2

The nearer ship makes the 45° angle. Find its distance, then subtract it from the distance of the farther ship.

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: AB is the lighthouse with $A B = 75$ m, and B is its top. D and C are the two ships on the same side of A, with C farther away. Let $A D = y$ and $D C = x$ .
The angles of depression equal the alternate angles at the ships: $∠ B C A = 3 0^{\circ}$ and $∠ B D A = 4 5^{\circ}$ .
1 mark
In right $△ B A C$ : $tan 3 0^{\circ} = \frac{A B}{A C} \Rightarrow \frac{1}{3} = \frac{75}{x + y}$
1 mark
$\Rightarrow x + y = 753$ …(i)
½ mark
In right $△ B A D$ : $tan 4 5^{\circ} = \frac{A B}{A D} \Rightarrow 1 = \frac{75}{y}$
1 mark
$\Rightarrow y = 75$ m …(ii)
½ mark
From (i) and (ii): $x = 753 - 75 = 75 (3 - 1) = 75 \times 0.732 = 54.9$ m.
So the distance between the two ships is 54.9 m.
1 mark

Answer: 54.9 m

Where the marks are

Correct labelled figure (1 mark)
Uses $tan 3 0^{\circ} = \frac{75}{x + y}$ (1 mark)
Gets $x + y = 753$ (½ mark)
Uses $tan 4 5^{\circ} = \frac{75}{y}$ (1 mark)
Gets $y = 75$ (½ mark)
Finds $x = 75 (3 - 1) = 54.9$ 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 paperSample paper 2024-25 (Basic): every question, by section.

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CBSE Class 10 Mathematics Basic sample paper 2024-25: question 35(B)

Hints

Step-by-step solution

Revise the chapter

More questions from Some Applications of Trigonometry