Question

The length of a shadow of a tower standing on a level plane is found to be 2y meters longer when the seen's altitude is 30° than when it was 45° prove that the height of the tower is y ( √3 + 1 ) meter.

Answer 1

In the right-angled triangle BCD.

tan 45° = `h/(BC)`

h = BC          ....(1)

In right-angled Δ ACD,

tan 30°  = `h/(2y + BC)`

⇒ `1/sqrt3 = h/(2y + h)`

⇒ `h(sqrt3 - 1) = 2y`

⇒ h = y ( √3 + 1 ) m
Hence proved.