Question

The length of the shadow of a tower standing on level plane is found to be '2y' metres longers when the sun's altitude is ${30}^{\circ }$ than when it is ${60}^{\circ }$. The height of the tower is m

• A. $3y$
• B. $\frac{\sqrt{3}y}{2}$
• C. $2\sqrt{3}y$
• D. $\sqrt{3}y$

Answer 1

The correct option is D

$\sqrt{3}y$

Let the height of the tower = h m
Let D be the point on the ground where the angle of elevation is ${30}^{\circ }$ and C be the point where the angle of elevation is ${60}^{\circ }$

$\therefore AC=x$ and $AD=x+2y$

In $\Delta ABC$

$tan{60}^{\circ }=\frac{h}{x}$

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

$x=\frac{h}{\sqrt{3}}\dots \left(i\right)$

In $\Delta ABD$

$tan30=\frac{h}{x+2y}$

$\frac{1}{\sqrt{3}}=\frac{h}{\frac{h}{\sqrt{3}}+2y}$

$\frac{1}{\sqrt{3}}=\frac{h}{\frac{h+2\sqrt{3}y}{\sqrt{3}}}$

$\frac{1}{\sqrt{3}}=\frac{h\sqrt{3}}{h+2\sqrt{3}y}$

$h+2\sqrt{3}y=3h$

$2\sqrt{3}y=2h$

$\therefore h=\sqrt{3}y$