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From the top of a lighthouse $70 m$ high with its base at sea level, the angle of depression of a boat is $15 °$ . The distance of the boat from the foot of the lighthouse is

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261.24 m

70 (2 - \sqrt{3}) m

265.25 m

70 (3 + \sqrt{3}) m

answer is A.

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Detailed Solution

It is given that from the top of a lighthouse which is

70 m

in height with its base at sea level, the angle of depression of a boat is found to be

15 °

. According to the given data the figure has been drawn below.
Question Image

Let AB=70m be the lighthouse and the boat is at point C.
Angle BCA is also equal to angle of depression =15

°

.
Now, in

Δ A B C,

tan θ = P e r p e n d i c u l a r b a s e

\tan (15 °) = \frac{AB}{BC}

,
We know that,

⇒ tan 1 5 ° = tan 60 ° − 45 ° ⇒ tan 1 5 ° = t a n 60 ° − t a n 45 ° 1 + t a n 60 ° t a n 45 ° ∵ tan (a − b) = tan a − tan b 1 + tan a tan b ⇒ tan 1 5 ° = 3 − 1 1 + 3 ∵ tan 60 ° = 3 and tan 45 ° = 1

Now,

3 − 1 3 + 1 = 70 B C

Now multiplying both numerator and denominator by

\sqrt{3} - 1

, we get,

3 − 1 3 + 1 × 3 − 1 3 − 1 = 70 B C

\Rightarrow \frac{(\sqrt{3} - 1) (\sqrt{3} - 1)}{(\sqrt{3})^{2} - 1}

\Rightarrow \frac{{(\sqrt{3} - 1)}^{2}}{3 - 1}

\Rightarrow \frac{3 + 1 - 2 \sqrt{3}}{2}

\Rightarrow \frac{2 (2 - \sqrt{3})}{2}

So,

BC = \frac{70}{(2 - \sqrt{3})} m

BC = \frac{70}{(2 - 1.732)}

∵ 3 = 1.732

BC = 261.24 m
Therefore, the distance of the boat from the foot of the lighthouse is 261.24 m.
Hence, Option 1 is correct.

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