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Question
A vertically straight tree, 15 m high, is broken by the wind in such a way that its top just touches the ground and makes an angle of 60° with the ground. At what height from the ground did the tree break?
Solution
Let AB be the tree of desired height x m and tree is broken by wind then tree makes an angle C = 60°. Let AC = 15 - x
Here we have to find height x
So we use trigonometric ratios.
In a triangle ACB
=> `sin C = (AB)/(AC)`
`=> sin 60^@ = x/(15 - x)`
`=> sqrt3/2 = x/(15 - x)`
`=> 15sqrt3 - sqrt3x = 2x`
`=> 15sqrt3 = 2x + sqrt3x`
`=> 15sqrt3 = x(2 + sqrt3)`
`=> x = (15sqrt3)/(2 +sqrt3)`
=> x = 6.9
Hence the height of tree is 6.9 m
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