Question

The angle of elevation of the top of a tower from two points at a distance of 5m. And 10m. from the base of the tower and in the same straight line with it are complementary. Prove that height of the tower is 5 root 2. ​

Answer 1

See attached figure ,

let AB be the tower

In triangle ABD,

tan theta = AB/BD

= tan theta = AB/5 ......................... (1)

In triangle ABC ,

tan (90 - theta) = AB/BC

= cot theta = AB / 10 [ ∵ Tan(90 - theta) = cot theta]

= tan theta = 10 / AB .......................... (2)

Equating (1) and (2) ,

$\frac{AB}{5}$ = $\frac{10}{AB}$

Cross multiply,

$AB^{2}$ = 5 x 10

$AB^{2}$  = 50

AB = $\sqrt{50}$

AB = 5$\sqrt{2}$

∴ PROVED