Question

I need help to solve this question​

Answer 1

See the figure at the top

let the height of tower AB be 'x'

and base AC be 'y' (when angle is 60°)

CD is the length of shadow increases = 40m

Now, in ∆BAC (see in fig.)

tan60° = perpendicular/base = x/y

√3 = x/y

y = x/√3

Now, in ∆BAD ( see in fig.)

tan30° = perpendicular/ base = x/(y+40)

1/√3 = x/(y+40)

y+40 = x√3-----eq1

put the value of y in equation 1

x/√3 + 40 = x√3

40 = x√3 - x/√3

40 = (3x - x)/√3

2x = 40√3

x = 20√3m

therefore, height of the tower is 20√3 m