• # question_answer 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.

 Let AB be the height of water level and CD be the height of hill Then, In $\Delta \,ABC$ $\Delta \,ABC$ $\tan \,30{}^\circ =\frac{10}{y}$ $y=10\sqrt{3}$                                         ?(i) In $\Delta \,ADE$ $\tan \,60{}^\circ =\frac{x}{y}$ $y=\frac{x}{\sqrt{3}}$                                           ?(ii) From (i) and (ii), we get $\frac{x}{\sqrt{3}}=10\sqrt{3}$ $x=10\times 3=30\,m$ $\therefore$ Distance of the hill from this ship is $10\sqrt{3}\,m$ and the height of the hill is $30+10=40\text{ }m$.