SSC CGL 7 Sep 2016 Paper

AB = Height of observer = 1.6 metre
CD = Height of tower = h metre
∴ DE = (h – 1.6) metre ; BC = AE = 45 metre
∠ DAE = 30°
∴ tan 30° = $\frac{\mathit{D}\mathit{E}}{\mathit{A}\mathit{E}}$
⇒ $\frac{1}{\sqrt{3}}$ = $\frac{\mathit{h}-1.6}{45}$
⇒ h - 1.6 = $\frac{45}{\sqrt{3}}$ = 15 $\sqrt{3}$
⇒ h - 1.6 = 15 × 1.732 = 25.98
⇒ h = (25.98 + 1.6) metre
= 27.58 metre