Question

From the top of a hill, the angles of depression of two consecutive kilometer stones due East are found to be ${30}^{\circ }$ and ${45}^{\circ }$ . Find the height of the hill.

Answer 1

Here, $AB$ is the hill.

Point $C$ and $D$ are the positions of the stones on the ground such that $CD=1km$

In $△ABC,\frac{AB}{BC}=\tan {45}^{\circ }$

$\frac{AB}{BC}=1$

$BC=AB.....\left(1\right)$

In $△ABD,\frac{AB}{BD}=\tan {30}^{\circ }$

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

$BC=\sqrt{3}AB....\left(2\right)$

$CD=1km$

$BD-BC=7km$

$\sqrt{3}AB-AB=1km$

$(\sqrt{3}-1)AB=1km$

$AB=\frac{1km}{\sqrt{3}-1}$

$=\frac{(\sqrt{3}+1)km}{(\sqrt{3}-1)(\sqrt{3}+1)}=\frac{(\sqrt{3}+1)km}{(\sqrt{3}{)}^2-1^2}$

$=\frac{(\sqrt{3}+1)km}{3-1}$

$=\frac{\sqrt{3}+1}{2}km$

Thus, the height of hill is $\frac{\sqrt{3}+1}{2}km$.