Question

From the top of a cliff 50 m high, the angles of depression of the top and bottom of a tower are observed to be 30° and 45°. The height of the tower is

(a) 50 m

(b) 50√3 m

(c) 50 (√3 - 1) m

(d) 50\(\big(1-\frac{\sqrt3}{3}\big)\)m

Answer 1

(d) 50\(\big(1-\frac{\sqrt3}{3}\big)\)m

Let the height of the cliff 

BD = 50 m and the height of the tower 

AE = h metres. 

Given, ∠DEC = 30°, ∠DAB = 45° Let BA = CE = x metres 

In rt. Δ DEC, 

tan 30° = \(\frac{DC}{CE}\) = \(\frac{50-h}{x}\)

⇒ \(\frac{1}{\sqrt3}\) = \(\frac{50-h}{x}\) ⇒ x = (50 - h)√3             ....(i)

In rt. Δ BAD, 

tan 45° = \(\frac{DB}{BA}\) ⇒ 1 = \(\frac{50}{x}\) ⇒ \(x\) = 50              ......(ii)

∴ From (i) and (ii) 

50 = (50 – h)√3 ⇒ 50 – 50√3 = −h√3

⇒ h = \(\frac{50(\sqrt3-1)}{\sqrt3}\) =  50\(\big(1-\frac{\sqrt3}{3}\big)\)m.