Let AB represent the tower and P the top of the cliff LP. If PX be the horizontal line through P, then ∠XPA = 30° and ∠XPB = 60°. Let the height of the tower be h metres. From A draw AM prep. to LP.
LM = AB = h
∴ MP = (200 – h) m
Again, ∠PBL = ∠XPB = 60°
∠PAM = ∠XPA = 30°
∴ From right ∆PLB, \(\frac{BL}{200}\) = cot 60° = \(\frac{1}{\sqrt3}\) ⇒ BL = \(\frac{200}{\sqrt3}\)
From right ∆PMA, \(\frac{AM}{MP}\) = cot 30°
∴ AM = MP cot 30° = (200 – h)√3
But AM = BL ∴ (200 – h) √3 = \(\frac{200}{\sqrt3}\) ⇒ (200 – h) × 3 = 200
⇒ 3h = 400 ⇒ h = \(\frac{400}{3}\) = 133\(\frac{1}{3}\)m.
