Question

From the top of a tower, the angle of depression of a point on the ground is 60°. If the distance of this point from the tower is \(\frac{1}{\sqrt3+1}\) metres, then the height of the tower is

(a) \(\frac{\sqrt3}{2}\)

(b) \(\frac{3\sqrt3}{2}\)

(c) \(\frac{3-\sqrt3}{2}\)

(d) \(\frac{3+\sqrt3}{2}\)

Answer 1

(c) \(\frac{3-\sqrt3}{2}\)

Let PQ be the tower and R be the point on the ground which is at a distance \(\frac{1}{\sqrt3+1}\) metres from the base of the tower, i.e., QR = \(\frac{1}{\sqrt3+1}\) metres.

Let the height of the tower PQ = h metres.

Given, angle of depression ∠APR = 60°, 

∴ ∠PRQ = ∠APR = 60° (AP || QR, alt ∠s) 

∴ In rt. Δ PQR, \(\frac{PQ}{QR}\) = tan 60°

⇒ \(\frac{h}{\frac{1}{\sqrt3+1}}\) = \(\sqrt3\) ⇒ (\(\sqrt3\) + 1) = \(\sqrt3\)

⇒ h = \(\frac{\sqrt3}{\sqrt3+1}\) x \(\frac{\sqrt3-1}{\sqrt3-1}\) = \(\frac{3-\sqrt3}{2}\) m.