Question

The length of the shadow of a tower standing on level ground is found to be 2x metres longer when the sun’s elevation is 30° than when it was 45°. The height of the tower in metres is

A. \((\sqrt3+1)x\)

B. \((\sqrt3-1)x\)

C. \(2\sqrt3x\)

D. \(3\sqrt2x\)

Answer 1

In Δ DBC 

tan 45° = h /y 

h = y (∵ tan 45° = 1)……….1 

In ΔACD 

tan 30° = \(\frac{h}{2x+y}\)

\(\frac{1}{\sqrt3}\) = \(\frac{h}{2x+y}\)

√3 h = 2x + h (from 1) 

2x = (√3 - 1) hh = \(\frac{2x}{(\sqrt3-1)}\)

h = (√3 +1)x meter