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 = 2500 m

∠BAD = 15° and ∠DAC = 45°

Let BD = x

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

Also DC = DQ + QC = 2500 + x

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

In ΔADC

$tan45=\frac{DC}{AD}$

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

$AD=5000+x$

$tan15=tan(45-30)=\frac{tan45-tan30}{1+tan45\times tan30}$

$=\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}$

$=\frac{\sqrt{3}-1}{\sqrt{3}+1}$

$In△ABD$

$tan45=\frac{BD}{AD}$

$=>\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{x}{5000+x}$

$(5000+x)(\sqrt{3}-1)=x(\sqrt{3}+1)$

$5000(\sqrt{3}-1)=2x$

$x=\frac{5000}{2}(\sqrt{3}-1)=2500(\sqrt{3}-1)$

Now BQ = BD + DQ = x+2500

$=2500(\sqrt{3}-1)+2500=2500\sqrt{3}$

=4330.12m
Hence the height of the cloud above the lake level is 4330.12m