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Angle of Depression

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Question

From the top of a lighthouse, the angles of depression of two ships on the opposite sides of it are observed to be $α$ and $β$ . If the height of the lighthouse be h metres and the line joining the ships passes through the foot of the lighthouse, find the distance between the ships is metres.

A

$\frac{h (c o s e c α + t a n β)}{t a n α t a n β}$

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B

$\frac{h (s e c α + t a n β)}{t a n α t a n β}$

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C

$\frac{h (t a n α + c o t β)}{t a n α t a n β}$

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D

$\frac{h (t a n α + t a n β)}{t a n α t a n β}$

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Solution

The correct option is D
$\frac{h (t a n α + t a n β)}{t a n α t a n β}$

Let AB be the lighthouse and C and D be the positions of the two ships. Then, AB = h metres.

Clearly, $∠ A C B = α$ and $∠ A D B = β$

Let AC = x metres and

AD = y metres

From right $Δ C A B$ , we have

$\frac{A C}{A B} = c o t α \Rightarrow \frac{x}{h} = c o t α \Rightarrow x = h c o t α$ ........... (i)

From right $Δ D A B$ , we have

$\frac{A D}{A B} = c o t β \Rightarrow \frac{y}{h} = c o t β \Rightarrow y = h c o t β$ ........... (ii)

Adding the corresponding sides of (i) and (ii), we get

$x + y = h (c o t α + c o t β) = h (\frac{1}{t a n α} + \frac{1}{t a n β})$

$\Rightarrow x + y = \frac{h (t a n α + t a n β)}{t a n α t a n β}$

Hence, the distance between the ships is $\frac{h (t a n α + t a n β)}{t a n α t a n β} m$

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