#### The angle of elevation of the top of a tower from certain point is 30°. If the observer moves 20 metres towards the tower, the angle of elevation of the top increases by 15°. Find the height of the tower.

Solution.

The angle of elevation of the top of a tower AB from certain point C is 30°
Let observer moves from C to D that is CD = 20m
Now angle of elevation increased by 15° that is 45° on point D
In $\bigtriangleup$ABD
$\tan 45^{\circ}= \frac{AB}{BD}$        $\left [ \because \tan \theta = \frac{Perpendicular}{Base} \right ]$
$1= \frac{AB}{BD}$
BD = AB        …(1)
In  $\bigtriangleup$ABC
$\tan 30^{\circ}= \frac{AB}{BC}$      $\left [ \because \tan \theta = \frac{Perpendicular}{Base} \right ]$
$\frac{1}{\sqrt{3}}= \frac{AB}{BD+DC}$      $\begin{bmatrix} \because \tan 30= \frac{1}{\sqrt{3}} & \\ BC= BD+DC & \end{bmatrix}$

$\frac{1}{\sqrt{3}}= \frac{AB}{BD+20}$        $\left ( \because DC= 20 \right )$
By cross multiplication we get
$BD+20= \sqrt{3}AB$
Now put the value of BD from equation (1) we have

$AB+20= \sqrt{3}AB$
$20= \sqrt{3}AB-AB$
$20= AB\left ( \sqrt{3}-1 \right )$
$AB= \frac{20}{\sqrt{3}-1}$
$AB= \frac{20}{1\cdot 732-1}= \frac{20}{0\cdot 732}$
AB = 27.322
Hence the height of the tower is 27.322 m