From the top of a light house, 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 light house be h meters and the line joining the ships passes through the foot of the light house, show that the distance between the ship is $\frac{h (t a n α + t a n β)}{t a n α t a n β}$ metres.

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Solution

Let AB be the light house and P and Q be the position of two ships .

Then ∠APB = α and ∠AQB = β

$I n △ A P B : t a n α = \frac{A B}{P B} = \frac{h}{P B}$

=> $P B = \frac{h}{t a n α}$ ----(1)

$I n △ A B Q : t a n β = \frac{A B}{B Q} = \frac{h}{B Q}$

=> $B Q = \frac{h}{t a n β}$ ----(2)

Now distance between the ships = PQ,

$I n △ A P B : t a n α = \frac{A B}{P B} = \frac{h}{P B}$

=> $P Q = P B + B Q = \frac{h}{t a n α} + \frac{h}{t a n β} = \frac{h t a n β + h t a n α}{t a n α t a n β} = \frac{h (t a n β + t a n α)}{t a n α t a n β} m$

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