Question

The angles of elevation of the top of a tower from two points at a distance of $4$ m and $9$ m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is $6$ m.

Answer 1

$\angle C$ and $\angle D$ are complementary .....given

In $△ABC,\frac{AB}{BC}=\tan Q$ ...(i)

In $△ABD,\frac{AB}{BD}=\tan (90-Q)$

$\frac{AB}{9}=\cot Q$ ...(ii)

$\frac{AB}{9}=\frac{1}{\tan Q}and\tan Q=\frac{7}{AB}$

Put value of $\tan Q$ in eqn (ii)

$\therefore \frac{AB}{BC}=\frac{9}{AB}$

$A{B}^2=9BC$

$AB=\sqrt{36}=6$

Height of tower $=6$m