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Question
A tree is broken by the wind. The top of that tree struck the ground at an angle of 30° and at a distance of 30. Find the height of the whole tree
Solution
Let AB represents the unbroken part and AC represent the broken part of the tree. The top of the tree (T) touches the ground at C.
BC = 30 cm, `angleACB = 30^@`
Total height of the tree = AB + AT = AB + AC .....(1)
In right angles `triangleABC`,
`tan angleACB = "AB"/"BC"`
`:. tan30^@ = "AB"/"BC"`
`:. 1/sqrt3 = (AB)/30`
`:. AB = 30/sqrt3 m` ......(2)
Also `cos angleACB = (BC)/(AC)`
`:. cos30^@ = (BC)/(AC)`
`:. sqrt3/3 = 30/(AC)`
`:.AC = 30 xx 2/sqrt3`
`:. AC = 60/sqrt3`
`:. AT = 60/sqrt3` ...(3)
Height of the tree = AB + AT ...[From (1)]
`= 30/sqrt3 + 60/sqrt3` ...[From (2) and (3)]
`=(30+60)/sqrt3 = 90/sqrt3 = 90/sqrt3 xx sqrt3/sqrt3 = (90sqrt3)/3`
∴ the height of the tree = `30sqrt3` m
= `30 xx 1.73 m` = 51.90 m
The height of the whole tree is 51.90 m.
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