Question

In the given figure, the top of a tower is observed from two points on the same horizontal line through a point on the base of the tower . If the angles of elevation at the two points be ${30}^{\circ }$ and ${45}^{\circ }$ , and the distance between them is $20$ m, find the height of the tower.

Answer 1

Let $AB$ be the height of the tower

$\tan {45}^o=\frac{AB}{BC}$

$AB=BC---\left(1\right)$

$\tan {30}^o=\frac{AB}{BD}$

$\frac{1}{\sqrt{3}}=\frac{AB}{BC+20}$

$BC+20=AB\sqrt{3}$

from equation $\left(1\right)$ $AB=BC$

$\therefore AB+20=AB\sqrt{3}$

$AB=\frac{20}{\sqrt{3}-1}\frac{\sqrt{3}+1}{\sqrt{3}+1}=10(\sqrt{3}+1)$

$AB=10(\sqrt{3}+1)m$

$\therefore$ height of the tower is $10(\sqrt{3}+1)m$