A vertically straight tree 15m high …

 
The height of the tree (DB) = 15 m
Suppose it broke at A and its top D touches the ground at C.

Suppose AB = h Then AD = AC = (15 - h) m
In {tex}\Delta ABC{/tex}
{tex}\sin {60^o} = \frac{{AB}}{{AC}}{/tex}
{tex} \Rightarrow \frac{{\sqrt 3 }}{2} = \frac{h}{{15 - h}}{/tex}
{tex} \Rightarrow 2h = 15\sqrt 3 - \sqrt 3 h{/tex}
{tex} \Rightarrow 2h + \sqrt 3 h = 15\sqrt 3 {/tex}
{tex} \Rightarrow h\left( {2 + \sqrt 3 } \right) = 15\sqrt 3 {/tex}
{tex} \Rightarrow h = \frac{{5\sqrt 3 }}{{2 + \sqrt 3 }}{/tex}
{tex} \Rightarrow h = \frac{{5\sqrt 3 }}{{2 + \sqrt 3 }} \times \frac{{2 - \sqrt 3 }}{{2 - \sqrt 3 }}{/tex}
{tex} \Rightarrow h = \frac{{30\sqrt 3 - 45}}{{4 - 3}}{/tex}
{tex} \Rightarrow h = 15\left( {2\sqrt 3 - 3} \right){/tex}
{tex} \Rightarrow h = 15\left[ {2 \times 1.73 - 3} \right]{/tex}
{tex} \Rightarrow h = 15\left[ {3.46 - 3} \right]{/tex}
{tex} \Rightarrow h = 15 \times 0.46{/tex}
{tex} \Rightarrow h = 6.9m{/tex}
{tex}\therefore{/tex} Height above the ground from where the tree broke is 6.9 meter.