Question

When the sun's altitude changes from ${30}^{\circ }$ to ${60}^{\circ }$, the length of the shadow of a tower decreases by $70m$. What is the height of the tower?

Answer 1

Let AD be the tower, BD be the initial shadow and CD be the final shadow.
Given that $BC=70m$, $\angle ABD={30}^{\circ }$ and $\angle ACD={60}^{\circ }$
Let CD = $x$ and AD = $h$.
From the right $\Delta CDA$,
$\tan {60}^{\circ }=\frac{AD}{CD}$
$\sqrt{3}=\frac{h}{x}$ $\dots \left(1\right)$
From the right $\Delta BDA$,
$\tan {30}^{\circ }=\frac{AD}{BD}$
$\frac{1}{\sqrt{3}}=\frac{h}{70+x}$ $\dots \left(2\right)$
Dividing (1) by (2)
$\frac{\sqrt{3}}{\frac{1}{\sqrt{3}}}=\frac{\frac{h}{x}}{\frac{h}{70+x}}$
$\Rightarrow 3=\frac{70+x}{x}$
$\Rightarrow 2x=70$
$\Rightarrow x=35$
Substituting this value of $x$ in (1), we have:-
$\sqrt{3}=\frac{h}{35}$
$\Rightarrow h=35\sqrt{3}=35\times 1.73=60.55\approx 60.6m$