Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30º and 45º respectively. If the lighthouse is 100 m high, the distance between the two ships is:

A. 173 m B. 200 m C. 273 m D. 300 m Answer: Option C
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Solution(By Apex Team)

Height and Distance solution image Let, BD be the lighthouse and A and C be the positions of the ships. Then, BD = 100 m, $\angle B A D=30^{\circ}, \angle B C D=45^{\circ}$ $\begin{array}{l} \tan 30^{\circ}=\frac{B D}{B A} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{100}{B A} \\ \Rightarrow B A=100 \sqrt{3} \\ \tan 45^{\circ}=\frac{B D}{B C} \\ \Rightarrow 1=\frac{100}{B C} \\ \Rightarrow B C=100 \end{array}$ Distance between the two ships; $\begin{array}{l} =A C \\ =B A+B C \\ =100 \sqrt{3}+100 \\ =100(\sqrt{3}+1) \\ =100(1.73+1) \\ =100 \times 2.73=273 m \end{array}$