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?

Answer: Option C

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
$$\begin{array}{rl}& \text{From}\text{the}\text{right}\Delta CDA\\ & \tan 60^{\circ }=\frac{AD}{CD}\\ & \sqrt{3}=\frac{h}{x}......\left(eq:1\right)\\ & \text{From}\text{the}\text{right}\Delta BDA\\ & \tan 30^{\circ }=\frac{AD}{BD}\\ & \frac{1}{\sqrt{3}}=\frac{h}{70+x}......\left(eq:2\right)\\ & \frac{eq:1}{eq:2}\Rightarrow \frac{\sqrt{3}}{\left(\frac{1}{\sqrt{3}}\right)}=\frac{\left(\frac{h}{x}\right)}{\left(\frac{h}{70+x}\right)}\\ & \Rightarrow 3=\frac{70+x}{x}\\ & \Rightarrow 2x=70\\ & \Rightarrow x=35\end{array}$$
Substituting this value of x in eq : 1, we have
$$\begin{array}{rl}& \sqrt{3}=\frac{h}{35}\\ & \Rightarrow h=35\sqrt{3}=35\times 1.73\\ & =60.55\approx 60.6\end{array}$$