Question

In the following figure, a tower AB is $20$ m high and its shadow BC on the ground is $20\sqrt{3}$ m long. Find the distance of the tip of the tower from point $C$.

Answer 1

Given tower $AB$ is $20$ m and $BC$ shadow on the ground is $20\sqrt{3}$ m

then in $\Delta ABC,$

$\tan \theta =\frac{AB}{BC}=\frac{20}{20\sqrt{3}}=\frac{1}{\sqrt{3}}=\tan {30}^0$

Let the distance $AC=x$

$\sin {30}^{\circ }=\frac{AB}{AC}=\frac{20}{x}\Rightarrow \frac{1}{2}=\frac{20}{x}\Rightarrow x=40\text{m}$

$\therefore$ The distance of the tip of the tower from the point C is $40\text{m}$