Question

A vertical tower stands on a horizontal plane and from a point on the ground at a distance of $30$ meters from the foot of the tower the angle of elevation ${60}^0$. The height of the tower is $(\sqrt{3}=1.73)$

Answer 1

From the adjoining figure the height of the tower MP is given y
$\frac{b}{30}=\tan {60}^0=\sqrt{3}=1.73$
$\therefore b=30\times \left(1.73\right)=51.9m$