Question

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

Answer 1

Here, AB is the hill and P and Q are two consecutive kilometer stones.

Let the height of the hill AB be $\text{h}$ m and BP = $x$ m.

PQ = 1 km = 1000 m

In ΔABP,

$tan{45}^o=\frac{AB}{BP}1=\frac{h}{x}x=h---\left(1\right)$

In ΔABQ,

$BQ=BP+PQ=x+1000tan{30}^o=\frac{AB}{BQ}\frac{1}{\sqrt{3}}=\frac{h}{x+1000}x+1000=\sqrt{3}h\sqrt{3}h=h+1000\text{[Using (1)]}(\sqrt{3}-1)h=1000h=\frac{1000}{(\sqrt{3}-1)}h=\frac{1000(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}h=\frac{1000(\sqrt{3}+1)}{(3-1)}h=\frac{1000(\sqrt{3}+1)}{2}h=500(\sqrt{3}+1)h=500(1.73+1)h=500\times 2.73=1365m$

Thus, the height of the hill is $1365m$