Question

From a point on the gound, the angle of elevation of the top of a tower is observed to be ${60}^{\circ }$. From a point 40 m vertically above the first point of observation, the angle of elevation of the top of the tower is ${30}^{\circ }$. Find the height of the tower and its horizontal distance from the point of observation.

Answer 1

We have, PQ as a vertical tower.

Now, in $\Delta YZQ$



$tan{30}^{\circ }=\frac{QZ}{YZ}$

$\frac{1}{\sqrt{3}}=\frac{QZ}{YZ}$

$YZ=QZ\sqrt{3}...\left(i\right)$

And, in $\Delta XPQ$

$tan{60}^{\circ }=\frac{QP}{XP}$

$\sqrt{3}=\frac{QZ+40}{XP}$

$YZ\sqrt{3}=QZ+40(\because XP=YZ)$

$QZ\sqrt{3}\left(\sqrt{3}\right)=QZ+40$ (using (i))

$3QZ=QZ+40$

$2QZ=40$

$QZ=20$

$\therefore$ Height of towr = $(40+20)m=60m$

and Horizontal distance = $QZ\sqrt{3}=20\sqrt{3}m$