From the top of a 45 m high light house, the angles of depression of two ships, on the opposite side of it, are observed to be 30° and 60°. If the line joining the ships passes through the foot of the light house, find the distance between the ships. (Use √(3) = 1.73)

MathsCBSE Class 10 · Standard · Board exam 2024, set 30/2/1 · 5 marks · Long answer

CBSE Class 10 Mathematics Standard 2024: question 34

Set in the CBSE Class 10 board exam of 2024, paper 30/2/1, section D, question 34, for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q34 · 5 marks · Long answerCompetency-basedSome Applications of Trigonometry

From the top of a $45$ m high light house, the angles of depression of two ships, on the opposite side of it, are observed to be $3 0^{\circ}$ and $6 0^{\circ}$ . If the line joining the ships passes through the foot of the light house, find the distance between the ships. (Use $3 = 1.73$ )

Source: official question paper from cbse.gov.in (PDF). See it in the full 2024 board paper, set 30/2/1.

Hints

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

Hint 1

Draw the lighthouse with one ship on each side; the angle of depression equals the angle of elevation from the ship.

Hint 2

Use tan in each right triangle with the 45 m side opposite the angle.

Hint 3

The required distance is the sum of the two horizontal distances.

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: $A B = 45$ m is the lighthouse with foot $A$ and top $B$ . Ships at $C$ and $D$ on opposite sides of $A$ , on a horizontal line through $A$ . Angles of depression from $B$ are $3 0^{\circ}$ (to $C$ ) and $6 0^{\circ}$ (to $D$ ), so $∠ B C A = 3 0^{\circ}$ , $∠ B D A = 6 0^{\circ}$ [alternate angles]. Let $C A = x$ , $A D = y$ .
1 mark
In right $△ B A D$ : $tan 6 0^{\circ} = \frac{A B}{A D} \Rightarrow 3 = \frac{45}{y}$
1 mark
$\Rightarrow y = \frac{45}{3} = 153$ m
½ mark
In right $△ B A C$ : $tan 3 0^{\circ} = \frac{A B}{C A} \Rightarrow \frac{1}{3} = \frac{45}{x}$
1 mark
$\Rightarrow x = 453$ m
½ mark
Distance between ships $= x + y = 453 + 153 = 603 = 60 \times 1.73 = 103.8$ m
1 mark

Answer: 103.8 m

Where the marks are

Correct figure (1 mark)
$tan 6 0^{\circ} = 45/ y$ (1 mark)
$y = 153$ (½ mark)
$tan 3 0^{\circ} = 45/ x$ (1 mark)
$x = 453$ (½ mark)
Distance $= 603 = 103.8$ 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 paper2024 board paper, set 30/2/1 (Standard): every question, by section.

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CBSE Class 10 Mathematics Standard 2024: question 34

Hints

Step-by-step solution

Revise the chapter

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