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Question
An aeroplane is flying at a height of 300 m above the ground. Flying at this height, the angles of depression from the aeroplane of two points on both banks of a river in opposite directions are 45° and 60° respectively. Find the width of the river. [Use `sqrt3` = 1⋅732]
Solution
Let CD be the height of the aeroplane above the river at some instant. Suppose A and B be two points on both banks of the river in opposite directions
Height of the aeroplane above the river, CD = 300 m
Now,
∠CAD = ∠ADX = 60º (Alternate angles)
∠CBD = ∠BDY = 45º (Alternate angles)
In right ∆ACD,
`tan 60^@ = (CD)/(AC)`
`=> sqrt3 = 300/(AC)`
`=> AC = 300/sqrt3 = 100sqrt3 m`
In right ∆BCD
`tan 45^@ = (CD)/(BC)`
`=> 1 = 300/(BC)`
=> BC = 300 m
∴ Width of the river, AB
= BC + AC
`= 300 + 100sqrt3`
`= 300 + 100 xx 1.73`
= 300 + 173
= 473 m
Thus, the width of the river is 473 m
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