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Question
A man is standing on the deck of a ship, which is 40 m above water level. He observes the angle of elevation of the top of a hill as 60° and the angle of depression of the base of the hill as 30°. Calculate the distance of the hill from the ship and the height of the hill `(sqrt(3) = 1.732)`
Solution
Let the height of the hill BE be h m and the distance of the hill from the ship be x m
In the right ∆ABD
tan 30° = `"AD"/"DB"`
`1/sqrt(3) = 40/x`
x = `40sqrt(3)` ...(1)
In the right ∆CDE
tan 60° = `"CE"/"DC"`
`sqrt(3) = ("h" - 40)/x`
x = `("h" - 40)/sqrt(3)` ...(2)
From (1) and (2) we get
`("h" - 40)/sqrt(3) = 40sqrt(3)`
h – 40 = 40 × 3
h = 120 + 40 = 160 m
Height of the hill = 160 m
Distance of the hill from the ship = `40 xx sqrt(3)`
= 40 × 1.732
= 69.28 m
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