Question

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

• A. $200\sqrt{3}$
• B. $400\sqrt{3}$
• C. $400$
• D. $100$

Answer 1

The correct option is C $400$


In $△PMC,$ we get
$\tan {30}^{\circ }=\frac{h-200}{x}\cdots \left(1\right)$
In $△PM{C}^{\prime },$ we get
$\tan {60}^{\circ }=\frac{h+200}{x}\cdots \left(2\right)$
From $\left(1\right)$ and $\left(2\right),$ we get
$\frac{h+200}{h-200}=3$
$\Rightarrow h+200=3h-600$
$\Rightarrow h=400$
Now, $\sin {30}^{\circ }=\frac{h-200}{PC}$
$\Rightarrow PC=2h-400$
$\therefore PC=400$ m