Question

A vertically straight 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 ${60}^o$ with the ground. At what height from the ground did the tree break?

Answer 1

Initially the Height of the Tree $=15m$

Now , when it breaks a part of it will remain Perpendicular to the Ground $\left(AB\right)$ and remaining part $\left(AD\right)$ will make angle of ${60}^o$

or we can say $AB+AD=15m.............\left(1\right)$

Now, in $△ABD$

$\sin {60}^o=\frac{AB}{AD}\Rightarrow AB=\frac{\sqrt{3}AD}{2}.............\left(2\right)$

From (1) and (2)

$AD\frac{\sqrt{3}+1}{2}=15m$

$\Rightarrow AD=\frac{30}{\sqrt{3}+1}=10.98m$ $(Take\sqrt{3}=1.73)$

From (1)

$AB+10.98=15\Rightarrow AB=4.02m$

So its broken at $4.02m$ from ground