The shadow of a tower standing on a level plane is found to be 50 m longer when Sun’s elevation is 30° than when it is 60°. Find the height of the tower.

Answers (1)

Solution.         According to question

     Let the height of tower = h
$\tan 60^{\circ}= \frac{h}{BD}$           $\left [ \because \tan \theta = \frac{Perpendicular}{Base} \right ]$
$\sqrt{3}= \frac{h}{BD}$                       $\left [ \because \tan 60^{\circ}= \sqrt{3} \right ]$
$BD= \frac{h}{\sqrt{3}}\cdots \left ( 1 \right )$
$\tan 30^{\circ}= \frac{h}{BC}= \frac{h}{BD+DC}$        $\left [ \because BC= BD+DC \right ]$
$\frac{1}{\sqrt{3}}= \frac{h}{BD+50}$
$BD+50= \sqrt{3h}$