Question

The length of a shadow of a tower standing on a level plane is found to be 2y meters longer when the sun’s altitude is 30° then when it was 45°. Prove that the height of the tower is `y(sqrt(3) + 1)` meters.

Answer 1

In the right-angled triangle BCD.

`tan 45^circ = h/(BC)`

h = BC   ....(1)

In right-angled ΔACD,

`tan 30^circ = h/(2y + BC)`

⇒ `1/sqrt(3) = h/(2y + h)`

⇒ `h(sqrt(3) - 1) = 2y`

⇒ `h = y(sqrt(3) + 1 ) m`

Hence proved.