NIFT UG 2013 Question Paper with solutions

Let AB be a tower of height = 30 m.
Two points on opposite sides are C and D.
∠ACB = 60°, ∠ADB = 45°
Distance BC = x m. and BD = y m
From ΔABD,
$\tan \left(45°\right)=\frac{\mathit{A}\mathit{B}}{\mathit{B}\mathit{D}}=\frac{30}{\mathit{y}}$
$1=\frac{30}{\mathit{y}}$
∴ BD = y = 30 m
From ΔABC,
$\tan \left(60°\right)=\frac{\mathit{A}\mathit{B}}{\mathit{B}\mathit{C}}=\frac{30}{\mathit{x}}$
$\sqrt{3}=\frac{30}{\mathit{x}}$
$\Rightarrow \mathit{x}=10\sqrt{3}\mathit{m}$
Distance between two points CD
= BC + BD
$=\mathit{x}+\mathit{y}$
$=\left(30+10\sqrt{3}\right)\mathit{m}$
$=47.32$m