The angle of elevation of a stationary cloud from a point $2500 \mathrm{~m}$ above a lake is $15^{\circ}$…
The angle of elevation of a stationary cloud from a point $2500 \mathrm{~m}$ above a lake is $15^{\circ}$ and from the same point the angle of depression of its reflection in the lake is $45^{\circ}$. The height (in metres) of the cloud above the lake, given that $\cot 15^{\circ}=2+\sqrt{3}$, is
2500
$2500 \sqrt{2}$
$2500 \sqrt{3}$
5000
Solution
In $\triangle E C D$
$\cot 15^{\circ}=\frac{E C}{C D}$
$\begin{array}{ll}\Rightarrow & E C=C D \cot 15^{\circ} \\ \Rightarrow & E C=C D(2+\sqrt{3}), \text { given }\end{array}$
In $\triangle E C F, \cot 45^{\circ}=\frac{E C}{C F}$
From Eq. (i)
$(H-h)(2+\sqrt{3})=H+h$
$\begin{array}{cc}\Rightarrow & H(2+\sqrt{3})-h(2+\sqrt{3})=H+h \\ \Rightarrow & H(1+\sqrt{3})=h(3+\sqrt{3}) \\ \Rightarrow & H=h\left(\frac{3+\sqrt{3}}{1+\sqrt{3}}\right)=2500\left(\frac{3+\sqrt{3}}{\sqrt{3}+1}\right) \\ \Rightarrow & H=2500 \sqrt{3} \mathrm{~m}\end{array}$
