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

Answer 1

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.