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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 $$\triangle ABC$$, 
$$\dfrac{AB}{BC} = \tan 45^o, BC=75$$m

In $$\triangle ABD$$, 
$$\dfrac{AB}{BD} = \tan 30^o$$

$$\Rightarrow \dfrac{75}{BC+CD}=\dfrac{1}{\sqrt{3}}$$

$$\Rightarrow \dfrac{75}{75+CD}=\dfrac{1}{\sqrt{3}}$$

$$\therefore CD=[75(\sqrt{3}-1)]$$m  

495463_466372_ans_f022fcea9d70471aa517100dd2e4ec45.png

Mathematics
RS Agarwal
Standard X

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