Question

The angle of elevation of a stationery cloud from a point $2500$ m above a lake is ${15}^{\circ }$ and the angle of depression of its reflection in the lake is ${45}^{\circ }$. What is the height of the cloud above the lake level? $(Use\tan {15}^{\circ }=0.268)$.

Answer 1

Let $A$ be the point of observation and $B$ be the position of the cloud and $PQ$ be the surface of lake. Then $C$ is the reflection of cloud and $AP=2500m$

Let $BD=x$

Then $DQ=AP=2500m$ and $QC=BQ=BD+DQ=2500+x$

Also $DC=DQ+QC=2500+x$

Also $DC=DQ+QC=2500+2500+x=5000+x$

$\Rightarrow$ In $△ADC,\tan {45}^o=\frac{DC}{AD}$

$\Rightarrow$ $1=\frac{5000+x}{AD}$

$\Rightarrow$ $AD=5000+x$ ------ ( 1 )

$\Rightarrow$ In $△ABD,\tan {15}^o=\frac{BD}{AD}$

$\Rightarrow$ $0.268=\frac{x}{5000+x}$

$\Rightarrow$ $0.268(5000+x)=x$

$\Rightarrow$ $1340+0.268x=x$

$\Rightarrow$ $0.732x=1340$

$\therefore$ $x=1830.60$

$\Rightarrow$ $BQ=BD+DQ=x+2500$

$\Rightarrow$ $BQ=1830.60+2500$

$\therefore$ $BQ=4330.6$

$\therefore$ The height of the cloud above the lake level is $4330.6m$