Question

# As observed from the top of a $$75$$ m high lighthouse from the sea-level, the angles of depression of two ships are $$\displaystyle { 30 }^{ \circ }$$ and $$\displaystyle { 45 }^{ \circ }$$. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

Solution

## In $$\triangle ABC$$, $$\dfrac{AB}{BC} = \tan 45^o, BC=75$$mIn $$\triangle ABD$$, $$\dfrac{AB}{BD} = \tan 30^o$$$$\Rightarrow \dfrac{75}{BC+CD}=\dfrac{1}{\sqrt{3}}$$$$\Rightarrow \dfrac{75}{75+CD}=\dfrac{1}{\sqrt{3}}$$$$\therefore CD=[75(\sqrt{3}-1)]$$m  MathematicsRS AgarwalStandard X

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