SSC CPO SI and ASI Model Paper 9

(4) Let FB and FD be the totalhorizontal and vertical distancesrespectively. After climbing till pointA, angle of elevation changes.

$\mathit{N}\mathit{o}\mathit{w},\mathit{t}\mathit{a}\mathit{n}30°=\mathit{A}\frac{\mathit{C}}{50}\Rightarrow \mathit{A}\mathit{C}=\frac{50}{\sqrt{3}}\mathit{m}$
$\mathit{t}\mathit{a}\mathit{n}45°=\frac{\mathit{D}\mathit{E}}{\mathit{A}\mathit{E}}=\frac{50\sqrt{3}-\mathit{E}\mathit{F}}{\mathit{A}}\mathit{E}$
$=\frac{50\sqrt{3}-\mathit{A}\mathit{C}}{\mathit{A}\mathit{E}}=\frac{50\sqrt{3}-\frac{50}{\sqrt{3}}}{\mathit{A}\mathit{E}}$
$\mathit{A}\mathit{E}=\frac{150}{\sqrt{3}}-\frac{50}{\sqrt{3}}=\frac{100\sqrt{3}}{3}\mathit{m}$
Thus, the total horizontal distance
is$\left[50+\frac{100\sqrt{3}}{3}\right]\mathit{m}$