height of a tree is 10m .It is bent by the wind in such a way that its top touches the ground and makes an angle of 60°with the ground .at what height from the bottom did the tree get bent
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Answered by
49
The total height of tree is 10 m
Then one part is x m
And, the other part will be (10-x)
The broken part forms the hypotenuse and the part firmed to the ground is the perpendicular.
sin60°=p/h
√3/2=x/10-x
10√3-√3x=2x
(10*1.73)-1.73x=2x
17.3=2x+1.73x
17.3=3.73x
1730/373=x
4.68=x
Then one part is x m
And, the other part will be (10-x)
The broken part forms the hypotenuse and the part firmed to the ground is the perpendicular.
sin60°=p/h
√3/2=x/10-x
10√3-√3x=2x
(10*1.73)-1.73x=2x
17.3=2x+1.73x
17.3=3.73x
1730/373=x
4.68=x
Answered by
15
let the tree be AB with its top A so
AB=10 m
now it bents and touch the ground so let that point be C making angle 90 with ground
now it became a triangle ABC
in triangle ABC angle B=60
now sin 60 = √3/2
also sin 60 =opp/Hypo
⇒√3/2=opp / 10
⇒opp=5√3
AB=10 m
now it bents and touch the ground so let that point be C making angle 90 with ground
now it became a triangle ABC
in triangle ABC angle B=60
now sin 60 = √3/2
also sin 60 =opp/Hypo
⇒√3/2=opp / 10
⇒opp=5√3
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