Question

On a straight line passing through the foot of a tower, two points C and D are at distances of 4 m and 16 m from the foot of tower respectively. If the angles of elevation from C and D of the top of the tower are complementary, then find the height of the tower.

Answer 1

Let height of tower AB = h m
Let $\angle ADB=\theta$
$\therefore \angle ACB={90}^{\circ }-\theta$



So, In $△$ ABC,

$\frac{AB}{BC}=tan(90-\theta )$

$\frac{h}{4}=cot\theta ...\left(i\right)$

In $△$ ABD,

$\frac{AB}{BD}=tan\theta$

$\frac{h}{16}=tan\theta$

Multiply eq. $\left(i\right)and\left(ii\right)$

$\frac{h}{4}\times \frac{h}{16}=cot\theta \times tan\theta$

$\frac{h^2}{64}=1$

$[\because cot\theta \times tan\theta =1]$

$\Rightarrow h^2=64\Rightarrow h=8m$

So, Height of tower = 8 metres.