When the sun's altitude changes from 30° to 60°, the length of the shadow of a tower decreases by 70m. What is the height of the tower?

Let AD be the tower, BD be the initial shadow and CD be the final shadow.
Given that BC = 70 m
∠ABD = 30°
∠ACD = 60°
Let CD = x, AD = h
From the right ⊿CDA
tan 60° = AD/CD
√3 = h/x ......eq(1)
From the right ⊿BDA
tan30° = AD/BD
1/√3 = h/(70 + x) ........ eq(2)
eq(1)/eq(2)
√3/(1/√3) = (h/x)/{h/(70+x)}
3 = 70 + x/x
2x = 70
x = 35
Substituting this value of x in eq:1, we have
√3 = h/35
h = 35√3
= 60.55 m