1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

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

Open in App
Solution

## 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

Suggest Corrections
30
Join BYJU'S Learning Program
Related Videos
Angle of Elevation
MATHEMATICS
Watch in App
Explore more
Join BYJU'S Learning Program