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Question
A person standing on the bank of a river observes that the angle of elevation of the top of a tree standing on the opposite bank is 60°. When he moves 40 m away from the bank, he finds the angle of elevation to be 30°. Find:
- the height of the tree, correct to 2 decimal places,
- the width of the river.
Solution
Let AB be the tree of height 'h' m and BC be the width of the river.
Let D be the point on the opposite bank of tree such that CD = 40 m.
Here ∠ADB = 30° and ∠ACB = 60°
Let speed of the boat be 'x' metre per minute.
In ΔABC,
`(AB)/(BC) = tan 60^circ = sqrt(3) `
`=> h/(BC) = sqrt(3)`
`=> h = BC sqrt(3)`
In ΔADB,
`(AB)/(BD) = tan 30^circ`
`=> h/(40 + BC) = 1/sqrt(3)`
`=> (BC sqrt(3))/(40 + BC) = 1/sqrt(3)`
`=> BC sqrt(3) * sqrt(3) = 40 + BC`
`=>` 3BC = 40 + BC
`=>` 3BC – BC = 40
`=>` 2BC = 40 m
`=> BC = 40/2 m`
`=>` BC = 20 m
∴ h = 20 × 1.732 = 34.64 m
Hence, height of the tree is 34.64 m and width of the river is 20 m.
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