Q28 / 75 · JEE Main 2020 (04 Sep Shift 2) · Mathematics

The angle of elevation of a cloud $C$ from a point $P, 200 m$ above a still take is $30^{o}$ . If the angle of depression of the image of $C$ in the lake from the point $P$ is $60^{o}$ , then $P C$ (in $m$ ) is equal to

Solution

$\tan 30 ° = \frac{x}{y} = \frac{1}{\sqrt{3}} \Rightarrow y = \sqrt{3} x and$

$\tan 60 ° = \frac{x + 400}{y}$

So, $x + 400 = \sqrt{3} y = 3 x$

$2 x = 400, x = 200$

$\sin 30 ° = \frac{200}{PC} \Rightarrow PC = 400$

Hence, the correct answer is option: 'C'.