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 $OP_{1}, \, OP_{2}, \, OP_{3}$ be the lengths of shadow.
In $\Delta AOP_{1}$ we have,
$tan \theta _{1} = \frac{O A}{O P_{1}} = \frac{h}{h} = 1$
$\Longrightarrow $ $\theta _{1}=\frac{\pi }{4}$
In $\Delta AOP_{2}$ we have,
$tan \theta _{2} = \frac{O A}{O P_{2}} = \frac{2}{2 h} = \frac{1}{2}$
$\Longrightarrow $ $\theta _{2}=tan^{- 1} \frac{1}{2}$
Similarly, $tan \theta _{3} = \frac{O A}{O P_{3}} = \frac{h}{3 h} = \frac{1}{3} \, $
$\theta _{3}=tan^{- 1} \frac{1}{3}$
$\therefore $ Sum of the angles
$= \theta _{1}+\theta _{2}+\theta _{3}$
$=\frac{\pi }{4}+tan^{- 1} \frac{1}{2} + tan^{- 1} ⁡ \frac{1}{3}$

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