SSC CGL 16 Aug 2015 Shift 2 Paper

PQ = Tower A = 45 metre
RS = Tower B = 15 metre,
QS = x metre (let)
∠ PSQ = 60° ; ∠ RQS = θ
From ΔPQS,
tan θ 60° = $\frac{\mathit{P}\mathit{Q}}{\mathit{R}\mathit{S}}$
⇒ $\sqrt{3}$ = $\frac{45}{\mathit{x}}$ ⇒ $\sqrt{3}$ x = 45
⇒ x = $\frac{45}{\sqrt{3}}$ = 15 $\sqrt{3}$ metre
From ΔRSQ
tan θ = $\frac{\mathit{R}\mathit{S}}{\mathit{Q}\mathit{S}}$ = $\frac{15}{15\sqrt{3}}$
⇒ tan θ = $\frac{1}{\sqrt{3}}$
⇒ tan θ = tan 30°
⇒ θ = 30°
∴ sin θ = sin 30° = $\frac{1}{2}$