# Solve this?

## From the foot of a column the angle of elevation of the top of a tower is 45° and from the top of column the angle of depression of the bottom of the tower is 30°. A man walks 10 meters from the bottom of the column towards the tower. He notices the angle of elevation of its top to be 60°. Find the heights of the column and the tower.

Apr 30, 2018

In $\Delta A B D , \angle A D B = {45}^{\circ}$

So the triangle is right isosceles

and $B D = A B = H = \text{ Height of the tower}$

Now $\frac{A B}{B X} = \tan {60}^{\circ}$

$\implies \frac{A B}{B D - D X} = \sqrt{3}$

$\implies \frac{H}{H - 10} = \sqrt{3}$

$\implies H = \left(H - 10\right) \sqrt{3}$

$\implies H = \frac{10 \sqrt{3}}{\sqrt{3} - 1} = \frac{10 \sqrt{3}}{2} \times \left(\sqrt{3} + 1\right) = 15 + 5 \sqrt{3}$m

Now $\frac{C D}{B D} = \tan {30}^{\circ}$

$\implies \frac{h}{H} = \frac{1}{\sqrt{3}}$

$\implies h = \frac{H}{\sqrt{3}} = \frac{15 + 5 \sqrt{3}}{\sqrt{3}} = 5 \sqrt{3} + 5$ m