When the sun’s altitude changes from  $30°\text{\hspace{0.17em}to\hspace{0.17em}}60°$, 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

$CD$ be the final shadow

Given that  $BC=70m,\angle ABD=30°,\angle ACD=60°$

Let $CD=x,AD=h$

In  $\Delta CDA$,

$\tan 60°=\frac{h}{x}$

$\Rightarrow \sqrt{3}=\frac{h}{x}\to \left(1\right)$

$\Rightarrow h=\sqrt{3}x$

In $\Delta ABD$,

$\tan 30°=\frac{h}{BD}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{BC+CD}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{70+x}\to \left(2\right)$

From (1) and (2)

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{\sqrt{3}x}{70+x}$

$\Rightarrow 3x=70+x$

$\Rightarrow 2x=70$

$\Rightarrow x=35$

Substituting this value of  $x$ in equation (1)

$h=\sqrt{3}\left(35\right)$

$h=35\sqrt{3}=35\times 1.73=60.55=60.6$