Question

A vertically staight 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 ${60}^{\circ }$ with the ground. At what height from the ground did the tree break?

Answer 1

The height of the three = 15 m
suppose it broke at 'C' and its top 'A' touches the ground at 'D'
Now, AC = CD, and angle BDC = 60°
BC = ?
Let BC = 'x'
So, AC = 15 - x and CD = 15 - x
In right angle BCD,
BC/CD = sin 60°
x/(15-x) = √3/2
2x = (15 -x) (√3)
2x = 15√3 - √3x
2x + √3x = 15√3
x(2 + √3) = 15√3
x = (15√3)/(2 + √3)
= {(15√3)/(2 + √3)} × {(2 - √3)/(2 - √3)}
= {(30√3) - (15 ×3)}/(4 - 3)
= {(30 × 1.73) - 45}/1
x = 51.9 - 45
x = 6.9 m
So, the height above the ground from the tree broke is 6.9 meter.