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Question
At a point A, 20 metres above the level of water in a lake, the angle of elevation of a cloud is 30˚. The angle of depression of the reflection of the cloud in the lake, at A is 60˚.
Find the distance of the cloud from A.
Solution
Let AB be the surface of the lake and P be the point of observation such that AP = 20 metres. Let C be the position of the cloud and C’ be its reflection in the lake.
Then CB = C’B. Let PM be perpendicular from P on CB.
Then m∠CPM=30º and m∠C'PM=60°
Let CM = h. Then CB = h + 20 and C’B = h + 20.
In ΔCMP we have,
`tan30^@="CM"/"PM"`
`1/sqrt3=h/"PM"`
`PM=sqrt3h....................(i)`
In ΔPMC' we have,
`tan 60^@="C'M"/"PM"`
`sqrt3="C'B+BM"/"PM"`
`sqrt3=(h+20+20)/"PM"................(ii)`
From equation (i) and (ii), we get
`sqrt3h=(h+20+20)/sqrt3`
3h=h+40
h=20m
Now,CB=CM + MB =h +20= 20+ 20 = 40.
Hence, the height of the cloud from the
surface of the lake is 40 metres.
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