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, show that the distance between the ships is $\frac{h (t a n α + t a n β)}{t a n α t a n β}$ metres. [3 MARKS]

Open in App

Solution

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

Alternative Method,

In $Δ D C B$

$t a n α = \frac{h}{y}$

$y = \frac{h}{t a n α}$ [1 MARK]

In $Δ D C A$

$t a n β = \frac{h}{x}$

$x = \frac{h}{t a n β}$ [1 MARK]

The distance between the two ships $= \frac{h}{t a n α} + \frac{h}{t a n β}$

$= h (\frac{t a n β + t a n α}{t a n β t a n α}) m$ [1 MARK]

Suggest Corrections

Similar questions

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.

Q. From the top light house, the angles of the depression of two ship on the opposite side of it are observed to be