Question

The angled of elevation of an aeroplane flying vertically above the ground as observed from two consecutive stones 1 km apart are 45° and 60°. The height of the aeroplane from the ground is

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

Answer 1

(d) $\frac{1}{2}(3+\sqrt{3})\mathrm{km}$
Let $AB$be the position of the aeroplane and $B$ and $C$ be the two stones (1 km apart) such that $BC=1$km.
Let $AD$ be the perpendicular meeting $BC$ produced at $D$. Thus, we have:
∠$ABD={45}^o$ and ∠$ACD={60}^o$
Let:
$AD=h$ km and $CD=x$ km


In ∆$ABD$, we have:

$\frac{AD}{BD}=\tan {45}^o=1$

⇒ $\frac{h}{(x+1)}=1$
⇒ $h=x+1$ (or) $x=h-1$ ...(i)

In ∆$ACD$, we have:

$\frac{AD}{CD}=\tan {60}^o=\sqrt{3}$

⇒ $\frac{h}{x}=\sqrt{3}$
⇒ $h=\sqrt{3}x$ (or) $x=\frac{h}{\sqrt{3}}$ ...(ii)
Putting the value of $x$ from (i) in (ii), we get:
$h-1=\frac{h}{\sqrt{3}}$
⇒ $\sqrt{3}(h-1)=h$
⇒ $h(\sqrt{3}-1)=\sqrt{3}$
⇒ $h=\frac{\sqrt{3}}{(\sqrt{3}-1)}$ $=\frac{\sqrt{3}}{(\sqrt{3}-1)}\times \frac{(\sqrt{3}+1)}{(\sqrt{3}+1)}=\frac{3+\sqrt{3}}{2}$ km
Hence, the height of the aeroplane from the ground is $\frac{1}{2}(3+\sqrt{3})$ km.