The angle of elevation of a cloud $\mathrm{C}$ from a point $\mathrm{P}, 200 \mathrm{~m}$ above a still lake is $30^{\circ}$. If the angle of depression of the image of $\mathrm{C}$ in the lake from the point $\mathrm{P}$ is $60^{\circ}$, then $\mathrm{PC}$ (in $\mathrm{m}$ ) is equal to :
Correct Option: 1
Let $\mathrm{PA}=\mathrm{x}$
For $\triangle \mathrm{APC}$
$\mathrm{AC}=\frac{\mathrm{PA}}{\sqrt{3}}=\frac{\mathrm{x}}{\sqrt{3}}$
$\mathrm{AC}^{1}=\mathrm{AB}+\mathrm{BCl}$
$\mathrm{AC}^{1}=\mathrm{AB}+\mathrm{BC}$
$\mathrm{AC}^{1}=400+\frac{\mathrm{x}}{\sqrt{3}}$
From $\Delta \mathrm{C}^{1} \mathrm{PA}: \mathrm{AC}^{1}=\sqrt{3} \mathrm{PA}$
$\Rightarrow\left(400+\frac{x}{\sqrt{3}}\right)=\sqrt{3} x \Rightarrow x=(200)(\sqrt{3})$
from $\Delta \mathrm{APC}: \mathrm{PC}=\frac{2 \mathrm{x}}{\sqrt{3}} \Rightarrow \mathrm{PC}=400$
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