Question

A tower is 60 m heigh. Its shadow is x metres shorter when the sun’s altitude is 45° than when it has been 30°, then x is equal to

Options

• 41.92 m

• 43.92 m

• 43 m

• 45.6 m

Answer 1

43.92 m

Explanation;

Hint:

In the right ∆ABC, tan 30° = `"AB"/"BC" = 60/(x + y)`

`1/sqrt(3) = 60/(x + y)`

⇒ x + y = `60sqrt(3)`

y = `60sqrt(3) - x`  ...(1)

In the right ∆ABD, tan 45° = `"AB"/"BD"`

1 = `60/y`

⇒ y = 60   ...(2)

From (1) and (2) we get

60 = `60sqrt(3) - x`

x = `60 sqrt(3) - 60`

= `60(sqrt(3) - 1)`

= 60(1.732 – 1)

= 60 × 0.732

x = 43.92 m