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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