Question

The shadow of a tower standing on a level ground is x metres long when the sun’s altitude is 30°, while it is y metres long when the altitude is 60°. If the height of the tower is \(45.\frac{\sqrt3}{2}\)m, then \(x\) – y is

(a) 45 m

(b) 45√3 m

(c) \(\frac{45}{\sqrt3}\)m

(d) 45.\(\frac{\sqrt3}{2}\)m

Answer 1

(a) 45 m

Let the given tower be AC

AC = \(\frac{45\sqrt3}{2}\)m.

When angle of elevation is 30°, length of shadow is \(x\) m. 

⇒ ∠ABC = 30°, BC = \(x\) When angle of elevation is 60°, length of shadow is y m. 

⇒ ∠ADC = 60°, DC = y 

In rt. ΔADC, \(\frac{AC}{DC}\) = tan 60°

⇒ \(\frac{\frac{45\sqrt3}{2}}{y}\) = √3 ⇒ y = \(\frac{45}{2}\)              ....(i)

In rt. Δ ABC, \(\frac{AC}{BC}\) = tan 30°

\(\frac{\frac{45\sqrt3}{2}}{x}\) = \(\frac{1}{\sqrt3}\) ⇒ \(x\) = \(\frac{135}{2}\)          .....(ii)

∴ x - y = \(\frac{135}{2}\) - \(\frac{45}{2}\) = \(\frac{90}{2}\) = 45 m.          (From (i) and (ii))