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?

Solution(By Examveda Team)

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