Question

The angle of elevation of the top of a tower at a distance of 120 m from a point A on the ground is ${45}^{\circ }$. If the angle of elevation of the top of a flagstaff fixed at the top of the tower, at A is ${60}^{\circ }$, then find the height of the flagstaff. [Use $\sqrt{3}=1.73$]

Answer 1

Let AB is the tower of height h meter and AC is flagstaff of height x meter.

$\angle APB={45}^o$ and $\angle BPC={60}^o$

$tan{60}^o=\frac{x+h}{120}$

$\sqrt{3}=\frac{x+h}{120}$

$x=120\sqrt{3}-h$

$tan{45}^o=\frac{h}{120}$

$1=\frac{h}{120}$

h = 120

Substitute the value of h in x,

$x=120\sqrt{3}-120$

$x=120(\sqrt{3}-1)$

$x=120(1.73-1)$

$x=87.6m$

Therefore the height of the flagstaff = $87.6m$