Question

The angle of elevation of the top of a vertical tower from a point on the ground is ${60}^0$ . from another point 10 m vertically above the first, its angle of elevation is${45}^0$ .Find the height of the tower.

Answer 1

R.E.F. Image.

Height of the tower, H = (h+10)m

In $△ABC,$

$tan{60}^{\circ }=\frac{h+10}{AB}$

$AB=\frac{h+10}{\sqrt{3}}\cdots \left(i\right)$

In $△DCO,$

$tan{45}^{\circ }=\frac{h}{AB}$

$\therefore AB=h\cdots \left(ii\right)$

comparing eq.(i) and (ii), we get :

$h=\frac{h+10}{\sqrt{3}}$

$\therefore \sqrt{3}h=h+10$

$\Rightarrow \sqrt{3}h-h=10$

$\Rightarrow h(\sqrt{3}-1)=10$

$\therefore h=\frac{10}{\sqrt{3}-1}$

Height of tower = h + 10

$=\frac{10}{\sqrt{3}-1}+10$

$=\frac{10+10\sqrt{3}-10}{\sqrt{3}-1}$

$H=\frac{10\sqrt{3}}{\sqrt{3}-1}m$