Question

The shadow of a tower standing on a level ground is found to be 60 m longer when the Sun’s altitude is ${30}^{\circ }$ than when it is ${60}^{\circ }$. Find the height of the tower.

• A. 30 m
• B. 20 m
• C. 30√3 m
• D. 20 m

Answer 1

The correct option is C

30√3 m

The given situation is represented aptly by the figure below

$tan{30}^{\circ }=\frac{DC}{AC}=\frac{DC}{AB+BC}\Rightarrow (AB+BC)=\frac{DC}{tan{30}^{\circ }}\dots \dots \left(1\right)Also,tan{60}^{\circ }=\frac{DC}{BC}\Rightarrow BC=\frac{DC}{tan{60}^{\circ }}\dots \dots \left(2\right)$
Subtracting eq(2) from (1)
$AB=\frac{DC}{tan{30}^{\circ }}-\frac{DC}{tan{60}^{\circ }}\Rightarrow 60=DC(\sqrt{3}-\frac{1}{\sqrt{3}})\Rightarrow DC=30\sqrt{3}m$