Question

An observer standing on a bank of river observes the top of a tree on the opposite bank making an angle of elevation ${60}^{\circ }$. He moves $30$ $m$ backward and observes the top of the tree making an angle of elevation ${30}^{\circ }$. Find the height of the tree and the width of the river.

Answer 1

Let $AB$ represent the tree of height, $h$ metre $BC$ be the width of the river, $xm.C$ and $D$ are the two different positions of the observer.
In right angle $△ABC$,
$\tan \angle ACB=\tan {60}^{\circ }=\frac{AB}{BC}$

$\therefore h=x\sqrt{3}$
$\therefore \frac{1}{\sqrt{3}}=\frac{h}{x+30}$
$\therefore x+30=h\sqrt{3}$
$\therefore h=\frac{x+30}{\sqrt{3}}$ .... (ii)
From equation (i) and (ii), we get
$x\sqrt{3}=\frac{x+30}{\sqrt{3}}$
$\therefore x\times \sqrt{3}\times \sqrt{3}=x+30$
$\therefore 3x=x+30$
$\therefore 2x=30$
$\therefore x=15$
$\therefore$ Width of the river $=15$ $m$.
Substituting the value of $x$ in equation (i)
$h=x\sqrt{3}$
$\therefore h=15\times \sqrt{3}$
$=15\times 1.73$
$=25.95$
$\therefore$ height of the tree $=25.95$ $m$.