As observed from the top of a lighthouse, the angles of depression of two ships are $30^{\circ}$ and $45^{\circ}$ . If one ship is exactly behind the other on the same side of the lighthouse and is 100m apart, find the height of the lighthouse from the sea-level.

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Solution

The given situation is represented in the figure above
$t a n 45^{\circ} = \frac{A B}{B C} \Rightarrow A B = B C t a n 60^{\circ} = B C \dots \dots (1) t a n 30^{\circ} = \frac{A B}{B D} = \frac{A B}{B C + C D} \Rightarrow A B = t a n 30^{\circ} (B C + C D) = \frac{(100 + B C)}{\sqrt{3}} \dots \dots (2) S u b t r a c t i n g e q (1) f r o m (2) \frac{100 + B C}{\sqrt{3}} - B C = 0 \Rightarrow 100 + B C - \sqrt{3} B C = 0 \Rightarrow 100 = (\sqrt{3} - 1) B C \Rightarrow B C = \frac{100}{\sqrt{3} - 1} = 50 (\sqrt{3} + 1) P u t t i n g t h i s i n e q (1) A B = 50 (\sqrt{3} + 1) m ∴ h e i g h t o f l i g h t h o u s e i s 50 (\sqrt{3} + 1) m$

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As observed from the top of a lighthouse, the angles of depression of two ships are 30∘ and 45∘. If one ship is exactly behind the other on the same side of the lighthouse and is 100m apart, find the height of the lighthouse from the sea-level.

As observed from the top of a lighthouse, the angles of depression of two ships are $30^{\circ}$ and $45^{\circ}$ . If one ship is exactly behind the other on the same side of the lighthouse and is 100m apart, find the height of the lighthouse from the sea-level.