Management Aptitude Test Dec 2019 Paper

Let $\mathrm{CD}$ be the altitude of the aeroplane and $\mathrm{A}$ and $\mathrm{B}$ be opposite points on the two banks of the river.
Then, $\mathrm{CD}=3000\mathrm{m},\angle \mathrm{DAC}=45°$ and $\angle \mathrm{DBC}=60°$In right ΔADC
$\tan 45°=\frac{\mathrm{DC}}{\mathrm{AD}}$
$\Rightarrow 1=\frac{3000}{\mathrm{AD}}\Rightarrow \mathrm{AD}=3000\mathrm{m}$
In right ΔBDC, $\tan 60°=\frac{\mathrm{CD}}{\mathrm{BD}}$
$\Rightarrow \sqrt{3}=\frac{3000}{\mathrm{BD}}$
$\Rightarrow \mathrm{BD}=\frac{3000}{\sqrt{3}}\mathrm{m}$
∴ Width of the river $=\mathrm{AD}+\mathrm{BD}$ $=\left(3000+\frac{3000}{\sqrt{3}}\right)\mathrm{m}$
$=3000+1000\sqrt{3}$ $=3000+1730=4730\mathrm{m}$