Question

the angle of elevation of a stationary cloud from a point 2500 m above a lake is 15°​

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=DCAD

1=5000+xAD

AD=5000+x

tan15=tan(45−30)=tan45−tan301+tan45×tan30

=1−1√31+1√3

=√3−1√3+1

In△ABD

tan45=BDAD

=>√3−1√3+1=x5000+x

(5000+x)(√3−1)=x(√3+1)

5000(√3−1)=2x

x=50002(√3−1)=2500(√3−1)

Now BQ = BD + DQ = x+2500

=2500(√3−1)+2500=2500√3

=4330.12m

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