Question

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

Answer 1

Let a man is standing on the deck of a ship at point a such that $AB=10m$

Let $CE$ be the hill

Thus, $AB=CD=10m$

The top and bottom of a hill are $E$ and $C$

Given:

The angle of depression of the base $C$ of the hill observed from $A$ is $30°$

The angle of elevation of the top $E$ of the hill observed from $A$ is $60°$

$\angle CAD=\angle BCA=30°$ ( alternate angles)

Let $AD=BC=xm$and $DE=\text{'}h\text{'}m$

In $∆ADE$ , $\tan 60°=$ Perpendicular / base

$\tan 60°=$$\frac{DE}{AD}$

$\sqrt{3}=\raisebox{1ex}{$h$}\!\left/ \!\raisebox{-1ex}{$x$}\right.$ $(\tan 60°=\sqrt{3})$

$h=\sqrt{3}x$………….equation $1$

In $∆ABC,$$\tan 30°=\frac{AB}{BC}$ $(\tan 30°=\frac{1}{\sqrt{3}})$

$\frac{1}{\sqrt{3}}=\frac{10}{x}$

$x=10\sqrt{3}m$ …………..equation $2$

Substitute the value of $x$ from equation ($2$) in equation ($1$),

we have, $h=\sqrt{3}x$$=\sqrt{3}\times 10\sqrt{3}=10\times 3=30m$

The height of the hill is $CE=CD+DE=10+30=40m$

Hence, the height of the hill is$40m$ & the distance of the hill from the ship is $10\sqrt{3m}$.