Question

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

Answer 1

Let AB be the tower and AB = x

Distance CD = 2y

In right ΔADB, we have

`tan theta = (AB)/(DB)`

`\implies tan 45^circ = x/(DB)`

`\implies 1 = x/(DB)`

`\implies` DB = x.

In right ΔACB, we have

`tan 30^circ = (AB)/(CB)`

`\implies 1/sqrt(3) = x/(CB)`

`\implies CB = sqrt(3)x`

`\implies x + 2y = sqrt(3)x`

∴ `sqrt(3)x - x = 2y`

`\implies x(sqrt(3) - 1) = 2y`

`x = (2y)/(sqrt(3) - 1)`

= `(2y(sqrt(3) + 1))/((sqrt(3) - 1)(sqrt(3) + 1))`

= `(2y(sqrt(3) + 1))/(3 - 1)`

= `(2y(sqrt(3) + 1))/2`

= `y(sqrt(3) + 1)`

∴ Required height of tower = `y(sqrt(3) + 1)`