Question

A man standing on the deck of a ship, which is 10 m above water level, observes the angle of elevation of the top of a hill as ${60}^{\circ }$ and the angle of depression of the base of hill as ${30}^{\circ }$. Find the distance of the hill from the ship and the height of the hill

Answer 1

Let AB be the height of water level and CD be the height of hill



Then,

In $\Delta ABC$

$tan{30}^{\circ }=\frac{10}{y}$

$y=10\sqrt{3}...\left(i\right)$

In $\Delta ADE$

$tan{60}^{\circ }=\frac{x}{y}$

$y=\frac{x}{\sqrt{3}}...\left(ii\right)$

From (i) and (ii), we get

$\frac{x}{\sqrt{3}}=10\sqrt{3}$

$x=10\times 3=30m$

$\therefore$ Distance of the hill from this ship is $10\sqrt{3}m$

and the height of the hill is $30+10=40m$.