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° with the ground. At what height from the ground did the tree break?

• A. $15(2\sqrt{3}-3)m$
• B. $15(2\sqrt{2}-2)m$
• C. $15(2\sqrt{5}-5)m$
• D. $15(3\sqrt{3}-2)m$

Answer 1

The correct option is A $15(2\sqrt{3}-3)m$


Let the initial height of tree H = 15 m =AB + AC
Let us assume that it is broken at A.
And the angle made by broken part with the ground ($\theta ={60}^{\circ }$)

Height from the ground to broken point
= h = AB
So, we have
H = AC + h
$\Rightarrow$ AC = (H - h) m
We get a right triangle formed by the above given data,
$\text{So,}sin\theta =\frac{\text{opposite side}}{\text{Hypotenuse}}$
$sin{60}^{\circ }=\frac{AB}{AC}$

$\frac{\sqrt{3}}{2}=\frac{h}{H-h}$

$\sqrt{3}(15-h)=2h$

$(2+\sqrt{3})h=15\sqrt{3}$

$h=\frac{15\sqrt{3}}{2+\sqrt{3}}\times \frac{2-\sqrt{3}}{2-\sqrt{3}}$

Rationalizing the denominator, we have
$h=\frac{\left(15\sqrt{3}\right)(2-\sqrt{3})}{2^2-(\sqrt{3}{)}^2}$

$h=15(2\sqrt{3}-3)$

Therefore, the height of broken point from the ground
$=15(2\sqrt{3}-3)$ m