Two ships are on either side of a 100 m lighthouse. The angles of elevation fro

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Two ships are on either side of a 100 m lighthouse. The angles of elevation from the two ships to the top of the lighthouse are 30° and 45°. Find the distance between the ships.

Options:

A . 300 m

B . 173 m

C . 273 m

D . 200 m

Answer: Option C
:
C

Let, BD be the lighthouse and A and C be the positions of the ships.
Then, BD = 100 m, $∠$ BAD = 30 $^{\circ}$ , $∠$ BCD = 45 $^{\circ}$
$t a n 30^{\circ} = \frac{B D}{B A} \Rightarrow \frac{1}{\sqrt{3}} = \frac{100}{B A} \Rightarrow B A = 100 \sqrt{3} t a n 45^{\circ} = \frac{B D}{B C} \Rightarrow 1 = \frac{100}{B C} \Rightarrow B C = 100$
Distance between the two ships
$= A C = B A + B C = 100 \sqrt{3} + 100 = 100 (\sqrt{3} + 1) = 100 (1.73 + 1) = 100 \times 2.73 = 273 m$

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