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 length of the shadow at ${30}^{\circ }$.

• A. 40 m
• B. 50 m
• C. 60 m
• D. 90 m

Answer 1

The correct option is D

90 m

In the above figure,

Let BD is the length of the shadow when sun is at 60° be x m

AB is the length of the shadow when sun is at 30° be (x + 60) m

AD = 60 m

In ΔABC,

$tan{30}^{\circ }=\frac{BC}{AB}$

$BC\sqrt{3}=AB$

i.e. $BC=\frac{AB}{\sqrt{3}}$= $\frac{60+x}{\sqrt{3}}$ -------(i)

In ΔBCD,

$tan{60}^{\circ }=\frac{BC}{BD}$

$BC=BD\sqrt{3}$

$BC=x\sqrt{3}$ ---------(ii)

Equating equation (i) and (ii), we get

$\Rightarrow$ $\frac{60+x}{\sqrt{3}}=x\sqrt{3}$

$\Rightarrow$ 2x = 60

$\Rightarrow$ x = 30

Hence, length of the shadow = (x + 60) m = 90 m