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Question
A man on the deck of a ship is 10 m above the water level. He observes that the angle of elevation of the top of a cliff is 45° and the angle of depression of the base is 300. Calculate the distance of the cliff from the ship and the height of the cliff.
Solution
Let H be the height of the cliff CE. And a man is standing on the ships at the height of 10 meters above from the water level.
Let AB = 10, BC = x, AD = BC, AB = DC, DE = h.
∠ACB = 30° and ∠DAE = 45°
We have tofind H and xT
The corresponding figure is as follows
In Δ ABC
`=> tan C = (AB)/(BC)`
`=> tan 30^@ = = 10/x`
`=> 1/sqrt3 = 10/x`
`=> x = 10sqrt3`
Again in ΔDAE
`=> tan A = (DE)/(AD)`
`=> tan 45^@ = h/x`
`=> 1 = h/x`
=> x = h
`=> x = 10sqrt3`
Therefore H = h + 10
`=> H = 10sqrt3 + 10`
`=> H = 10(sqrt3 + 1)`
`=> H = 27.32`
Hence the required distance is `10sqrt3` and height is 27.32 m
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