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?

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#### 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

Concept: Heights and Distances

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