Question

The length of the shadows of a vertical pole of height $h$ , thrown by the sun's rays at three different moments are $h,2h$ and $3h$ . The sum of the angles of elevation of the rays at these three moments is equal to

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{4}$
• D. $\frac{\pi }{6}$

Answer 1

Answer: (A)

Let $OA$ be the vertical pole of height $h$ and $O{P}_1,O{P}_2,O{P}_3$ be the lengths of shadow.
In $\Delta AO{P}_1$ we have,
$tan{\theta }_1=\frac{OA}{O{P}_1}=\frac{h}{h}=1$
$⟹$ ${\theta }_1=\frac{\pi }{4}$
In $\Delta AO{P}_2$ we have,
$tan{\theta }_2=\frac{OA}{O{P}_2}=\frac{2}{2h}=\frac{1}{2}$
$⟹$ ${\theta }_2=ta{n}^{-1}\frac{1}{2}$
Similarly, $tan{\theta }_3=\frac{OA}{O{P}_3}=\frac{h}{3h}=\frac{1}{3}$
${\theta }_3=ta{n}^{-1}\frac{1}{3}$
$\therefore$ Sum of the angles
$={\theta }_1+{\theta }_2+{\theta }_3$
$=\frac{\pi }{4}+ta{n}^{-1}\frac{1}{2}+ta{n}^{-1}\frac{1}{3}$

$=\frac{\pi }{4}+{\left(tan\right)}^{-1}\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}\times \frac{1}{3}}\right)$
$=\frac{\pi }{4}+ta{n}^{-1}1$
$=\frac{\pi }{4}+\frac{\pi }{4}=\frac{\pi }{2}$