Q. 504.3( 10 Votes )

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)

Answer :

In the fig B is the position of the cloud and C is the point of reflection of the cloud in the lake.



In the fig BD = x


DQ = AP= 2500m


QC = BQ = BD + DQ = (2500 + x)m


DC = DQ +QC = 2500 + 2500 + x = (5000 + x)m


In ∆QRT

tan 45° =


1 =


AD = 5000 + x …………….(1)


Tan15° = tan (45° - 30°) = [Using formula tan(α-β) = ]


Tan15° = =


In ∆ABD


tan 15° =


=


(√3+1)x = (√3+1)(5000+x)


√3x + x = 5000√3 + √3x – 5000 - x


2x = 5000(√3-1)


x = 2500(√3-1)m


Now BQ = BD + DQ = x + 2500


(On substituting value of x)


BQ = 2500(√3-1) + 2500


2500√3 – 2500 + 2500


2500√3m


Therefore height of cloud is 2500√3m

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