SSC CGL 26 Oct 2014 Shift 2 Paper

AC = Flag
AB = building = 10 metre
∠ APB = 30°; ∠ CPB = 45°
In Δ APB,
tan 30° = $\frac{\mathit{A}\mathit{B}}{\mathit{P}\mathit{B}}$
⇒ $\frac{1}{\sqrt{3}}$ = $\frac{10}{\mathit{P}\mathit{B}}$
⇒ PB = $10\sqrt{3}$ metre
In D PBC,
tan 45° = $\frac{\mathit{B}\mathit{C}}{\mathit{P}\mathit{B}}$
⇒ 1 = $\frac{\mathit{A}\mathit{B}+\mathit{A}\mathit{C}}{\mathit{P}\mathit{B}}$
⇒ PB = AB + AC
⇒ 10 $\sqrt{3}$ = 10 + AC
⇒ AC = 10 $\sqrt{3}$ - 10
= 10 ($\sqrt{3}$ - 1) metre
= 10 (1.732 – 1) metre
= 10 × 0. 732 = 7.32 metre