A 15m high tree is broken 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 is the tree broken?

Given:​

Height of tree = 15 m 

After broke , the angle between ground and tree = 60⁰ 

Formula Used: 

​$$\sin 60^\circ = \frac{Perpendicular}{Hypotenuse} = \frac{\sqrt3}{2}$$ 

Solution:​

​$$\sin 60^\circ = \frac{Perpendicular}{Hypotenuse} = \frac{\sqrt3}{2} \\ \ \\\frac{15 -x}{x} = \frac{\sqrt3}{2} \\ \implies 30 - 2x = \sqrt3x \\ \sqrt3x+2x = 30 \\ x = \frac{30}{2+\sqrt3}$$   

Broken height = $$15 - \frac{30}{2+\sqrt3} =\bf \frac{15\sqrt3}{2+\sqrt3} \, m$$ ​​​