Question : A Navy captain is going away from a lighthouse at the speed of $4(\sqrt3–1)$ m/s. He observed that it took him 1 minute to change the angle of elevation of the top of the lighthouse from 60° to 45°. What is the height (in meters) of the lighthouse?
Option 1: $240\sqrt{3}$
Option 2: $480(\sqrt{3}–1)$
Option 3: $360\sqrt{3}$
Option 4: $280\sqrt{2}$
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Correct Answer: $240\sqrt{3}$
Solution : Given: A Navy captain is going away from a lighthouse at the speed of $4(\sqrt3-1)$m/s. Let the base length be $B_{1}$ when the elevation is 60° and the base length be $B_{2}$ when the elevation is 45° and the height of the lighthouse be H. Distance between the $B_{2} - B_{1} = 4[(\sqrt{3} - 1)] × 60$ metres. When the elevation is 60°, then $\tan \ \theta = \frac{H}{B_{1}}$ $⇒\tan \ 60° = \frac{H}{B_{1}}$ $⇒\sqrt{3} = \frac{H}{B_{1}}$ $⇒B_{1} = \frac{H}{\sqrt{3}}$ When the elevation is 45°, then $\tan \ \theta = \frac{H}{B_{2}}$ $⇒\tan \ 45° = \frac{H}{B_{2}}$ $⇒B_{2} = H$ Now, $B_{2} - B_{1} = 4[(\sqrt{3} - 1)] × 60$ metres $⇒H - \frac{H}{\sqrt{3}}= 4[(\sqrt{3} - 1)] × 60$ $⇒H(\frac{\sqrt{3} \:-\: 1}{\sqrt{3}}) = 4[(\sqrt{3} - 1)] × 60$ $\therefore H = 240\sqrt{3}$ m Hence, the correct answer is $240\sqrt{3}$.
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