Question

The angle of elevation of the top of a hill from a point on the horizontal plane passing through the foot of the hill is found to be ${45}^{\circ }$. After walking a distance of $80$ meters towards the top, up a slope inclined at an angle of ${30}^{\circ }$ to the horizontal plane, the angle of elevation of the top of the hill becomes ${75}^{\circ }$. Then the height of the hill (in meters) is

Answer 1

$x=80\cos {30}^{\circ }=40\sqrt{3}$
$y=80\sin {30}^{\circ }=40$
In $\Delta$ ADC
$\tan {45}^{\circ }=\frac{h}{x+z}\Rightarrow h=x+z$
$\Rightarrow h=40\sqrt{3}+z\cdots \left(1\right)$
In $\Delta$ EDF
$\tan {75}^{\circ }=\frac{h-y}{z}$
$2+\sqrt{3}=\frac{h-40}{z}\Rightarrow z=\frac{h-40}{2+\sqrt{3}}\cdots \left(2\right)$
Put the value of $z$ from $\left(1\right)$
$h-40\sqrt{3}=\frac{h-40}{2+\sqrt{3}}$
$h(1+\sqrt{3})=40(2\sqrt{3}+3-1)$
$h(1+\sqrt{3})=80(1+\sqrt{3})$
$h=80$