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As observed from the top of light house, 50 m above sea level the angle of depression of ship, sailing directly towards it, changes from $30 °$ to $45 °$ . Determine the distance travelled by ship.

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400 (1 − 3) m

50 (1 − 3) m

200 (1 − 3) m

100 (1 − 3) m

answer is C.

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Detailed Solution

Given that, as observed from the top of lighthouse, 50m above sea level, the angle of depression of ship, sailing directly towards it, changes from

30 °

45 ° .

We have to find the distance traveled by ship.
In a right-angle triangle,

tan θ = Opposite side Adjacent side .

Suppose the initial and final position of the ship be A and B as d be the distance travelled be the ship.
Then, AB = d.
Question Image

Let the position of the observer is at A. Then O P=100 m.
From the right angle triangle OPB,

tan45 ° = OP BP ⇒ 1= 100 BP ⇒ BP=100 m

From the right angle triangle OPA,

tan 30 ° = OP AP ⇒ 13 = 100 d+BP ⇒ d+100=100 3 ⇒ d=100-100 3

The distance travelled by ship is

100 (1 − 3) m

.
Therefore, the correct option is 3.

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