#### The angle of elevation of the top of a vertical tower from a point on the ground is 60°. From another point, 10 m vertically above the first, its angle of elevation is 45°. Find the height of the tower.

Solution.         According to question

Let h is the height of the tower
In $\bigtriangleup$ABE
$\tan \theta = \frac{P}{B}$
$\tan 60^{\circ} = \frac{h}{AB}$             $\left ( \because \theta = 60^{\circ} \right )$
$\sqrt{3}= \frac{h}{AB}$       $\left ( \because \tan 60^{\circ}= \sqrt{3} \right )$
$AB= \frac{h}{\sqrt{3}}\: \cdots \left ( 1 \right )$
In $\bigtriangleup$EDC
$\tan \theta = \frac{P}{B}$
$\tan 45^{\circ} = \frac{h-10}{DC}$
$1= \frac{h-10}{AB}$    $\left ( \because AB-DC,\tan 45^{\circ}= 1 \right )$
$AB= h-10\: \cdots \left ( 2 \right )$
from equation (1) and (2)
$h-10= \frac{h}{\sqrt{3}}$
$\sqrt{3}h-\sqrt{3}10= h$
$\sqrt{3}h-h= \sqrt{3}\times 10$
$h\left ( \sqrt{3}-1 \right )= 10\sqrt{3}$
$h= \frac{10\sqrt{3}}{\sqrt{3}-1}$
Hence the height of the tower
$h= \frac{10\sqrt{3}}{\sqrt{3}-1}m$