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As observed from the top of a lighthouse, $100 m$ above sea level, the angle of depression of a ship, sailing directly towards it, changes from $30 °$ to $45 ° .$ ,the distance travelled by the ship during the period of observation is

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43.2 m

63.2 m

53.2 m

73.2 m

answer is D.

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Detailed Solution

Given that the top of a lighthouse is

100 m

above sea level, the angle of depression of a ship, sailing directly towards it, changes from

30 °

45 ° .

Consider the figure,
Question Image

We know,

tanθ = \frac{Perpendicular}{Base}

\tan 30 ° = \frac{1}{\sqrt{3}}

\tan 45 ° = 1

Now in triangle ABC,

\tan 45 ° = \frac{AB}{BC}

\Rightarrow \tan 45 ° = \frac{100}{CB}

\Rightarrow CB = \frac{100}{\tan 45 °}

In triangle ADB,

\tan 30 ° = \frac{AB}{BD}

\Rightarrow \tan 30 ° = \frac{100}{BD}

\Rightarrow BD = \frac{100}{\tan 30 °}

The distance travelled by the ship,
BD-CB
On substituting values we get,

d = 100 tan 30 ° − 100 tan 45 ° ⇒ d = 1003 − 100 ⇒ d = 73.2 m

The distance travelled by the ship during the period of observation is 73.2 m.
Therefore, option 4 is correct.

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