Question

A tower is 120 m high. Its shadow is x m shorter, when the sun’s altitude is 60° than when it was 45°. Find x correct to nearest metre.

Answer 1

Let AB be the tower and its shadows be BD and BC corresponding to the angle of elevations 60° and 45° respectively.

In Δ ABC, tan 45° = \(\frac{AB}{BC}\)

⇒ 1 = \(\frac{120}{BC}\) ⇒ BC = 120m

In Δ ABD, tan 60° = \(\frac{AB}{BD}\) ⇒ √3 = \(\frac{120}{BD}\)

⇒ BD = \(\frac{120}{\sqrt3} = \frac{120}{\sqrt3}\times\frac{\sqrt3}{\sqrt3}=\frac{120\sqrt3}{3}=40\sqrt3\) metres.

∴ x = BC - BD = 120 - 40√3 = 120m – 69.28 m 

= 50.72 m = 51 m ( to the nearest m).