Question

From the top of a hill, the angles of depression of two consecutive km stones due east are found to be 30° and 45°. The height of the hill is

(a) $\frac{1}{2}(\sqrt{3}-1)\mathrm{km}$
(b) $\frac{1}{2}(\sqrt{3}+1)\mathrm{km}$
(c) $(\sqrt{3}-1)\mathrm{km}$
(d) $(\sqrt{3}+1)\mathrm{km}$

Answer 1

(b) $\frac{1}{2}(\sqrt{3}+1)\mathrm{km}$
Let $AB$ be the hill making angles of depression at points $C$ and $D$ such that ∠$ADB={45}^o$, ∠$ACB={30}^o$ and $CD=1$ km.
Let:
$AB=h$ km and $AD=x$ km

In ∆$ADB$, we have:
$\frac{AB}{AD}=\tan {45}^o=1$

⇒ $\frac{h}{x}=1$ ⇒ $h=x$ ...(i)
In ∆$ACB$, we have:
$\frac{AB}{AC}=\tan {30}^o=\frac{1}{\sqrt{3}}$

⇒ $\frac{h}{x+1}=\frac{1}{\sqrt{3}}$ ...(ii)
On putting the value of $h$ taken from (i) in (ii), we get:
$\frac{h}{h+1}=\frac{1}{\sqrt{3}}$
⇒ $\sqrt{3}h=h+1$
⇒ $(\sqrt{3}-1)h=1$
⇒ $h=\frac{1}{(\sqrt{3}-1)}$
On multiplying the numerator and denominator of the above equation by $(\sqrt{3}+1)$, we get:
$h=\frac{1}{(\sqrt{3}-1)}\times \frac{(\sqrt{3}+1)}{(\sqrt{3}+1)}=\frac{(\sqrt{3}+1)}{3-1}=\frac{(\sqrt{3}+1)}{2}$ $=\frac{1}{2}(\sqrt{3}+1)$ km

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