A trees 12 m high is …

Let AB be the tree of height 12 metres. Suppose the tree is broken by the wind at point C and the part CB assumes the position CO and meets the ground at O.
Let AC = x. Then, CO = CB = 1 2 - x. It is given that {tex}\angle A O C{/tex} = 60^o
In {tex}\triangle O A C,{/tex} we have

{tex}\sin 60 ^ { \circ } = \frac { A C } { O C }{/tex}
{tex}\Rightarrow \quad \frac { \sqrt { 3 } } { 2 } = \frac { x } { 12 - x }{/tex}
{tex}\Rightarrow \quad 12 \sqrt { 3 } - \sqrt { 3 } x = 2 x{/tex}
{tex}\Rightarrow \quad 12 \sqrt { 3 } = x ( 2 + \sqrt { 3 } ){/tex}
{tex}\Rightarrow \quad x = \frac { 12 \sqrt { 3 } } { 2 + \sqrt { 3 } } = \frac { 12 \sqrt { 3 } } { 2 + \sqrt { 3 } } \times \frac { 2 - \sqrt { 3 } } { 2 - \sqrt { 3 } } = 12 \sqrt { 3 } ( 2 - \sqrt { 3 } ){/tex}
{tex}\Rightarrow \quad x = 24 \sqrt { 3 } - 36 = 5.569{/tex} metres
Hence, the tree is broken at a height of 5.569 metres from the ground.