Question

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

(a) $\frac{1}{2}(\sqrt{3}-1)$ km

(b) $\frac{1}{2}(\sqrt{3}+1)$ km

(c) $(\sqrt{3}-1)$ km

(d) $(\sqrt{3}+1)$ km

Answer 1

Let AB be the hill and C and D be the points such that $\angle ADB={45}^o,\angle ACB={30}^o$ and CD = 1 km.

Let AB = h m and AD = x km. Then,

$\frac{AB}{AD}=tan{45}^o=1\Rightarrow \frac{h}{x}=1\Rightarrow h=x---\left(1\right)]\frac{AB}{AC}=tan{30}^o=\frac{1}{\sqrt{3}}\Rightarrow \frac{h}{x+1}=\frac{1}{\sqrt{3}}\text{Put x = h from (1) in here. We get,}\frac{h}{h+1}=\frac{1}{\sqrt{3}}\Rightarrow \sqrt{3}h=h+1\Rightarrow (\sqrt{3}-1)h=1\Rightarrow h=\frac{1}{\sqrt{3}-1}=\frac{1(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}=\frac{(\sqrt{3}+1)}{3-1}=\frac{1}{2}(\sqrt{3}+1)$

The height of the hill is $\frac{1}{2}(\sqrt{3}+1)$