A shadow of a tower standing on level ground is found to be 40√3 meters longer when the Sun's altitude is 30° than when it is 60°. The height of the tower is:

Explanation

Let the height of the tower be h i.e. h = AB In △ABC, tan 60° = AB/BC → √3 = AB/BC → AB = BC√3 Now, In △ABD, tan 30° = AB/BD → 1/√3 = BC√3 / (BC + CD) → 3 BC = BC + CD → 3 BC - BC = CD → 2 BC = 40/√3 → BC = 20/√3 Now, The height of the tower = AB = BC√3 = 20/√3 × √3 = 60 m ∴ The height of the tower is 60 meters.