Question

A man on the deck of a ship is $10m$ above the water level. He observes that the angle of elevation of the top of a cliff is ${60}^{\circ }$ and the angle of depression of the base is ${30}^{\circ }$. Calculate the distance of the cliff from the ship and the height of the cliff.

Answer 1

Let $AB$ be the deck of the ship. $AB=10m$

Suppose $CE$ be the cliff. $C$ and $E$ are the top and bottom of the cliff.

Let $CE=hm$

$CD=AB=10m$

$\therefore$ $DE=CE-CD=(h-10)m$

$\Rightarrow$ In $△ADE,\tan {45}^o=\frac{DE}{AD}$

$\Rightarrow$ $1=\frac{h-10}{AD}$

$\therefore$ $AD=(h-10)m$ ------- ( 1 )

$\Rightarrow$ In $△ADC,\tan {30}^o=\frac{CD}{AD}$

$\Rightarrow$ $\frac{1}{\sqrt{3}}=\frac{10}{AD}$

$\therefore$ $AD=10\sqrt{3}m$ ------ ( 2 )

( From 1 and 2)

$\Rightarrow$ $h-10=10\sqrt{3}$

$\Rightarrow$ $h=10(\sqrt{3}+1)=10(1.732+1)=27.32m$

$\Rightarrow$ Height of the cliff is $27.32m$ and distance of cliff from the ship $BC=AD=17.3m$