Question

A pole $5m$ high is fixed on the top of a tower. The angle of elevation of the top the pole observed from a point $A$ on the ground is ${60}^{\circ }$ and the angle of depression of the point $A$ from the top of the tower is ${45}^{\circ }.$ Find the height of the tower.

Answer 1

REF.Image

$tan{60}^{\circ }=\frac{BD}{AB}$ $tan{45}^{\circ }=\frac{BC}{AB}$

$\sqrt{3}=\frac{BD}{AB}$ $1=\frac{BC}{AB}$

$AB=\frac{BD}{\sqrt{3}}$ _____ (1) $AB=BC$ _____ (2)

From (1) & (2)

$\frac{BD}{\sqrt{3}}=BC$

$(BC+5)=\sqrt{3}BC$

$5=1.732BC-BC$

$5=0.732BC$

$BC=\frac{5}{0.732}$

$BC=6.83m$