Question

A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle ${30}^{\circ }$ with it. The distance between the foot of the tree to the point where the top touches the ground is 9 m. Find the height of the tree.

• A. $4\sqrt{3}metres$
• B. $5\sqrt{3}metres$
• C. $2\sqrt{3}metres$
• D. $9\sqrt{3}metres$
  

Answer 1

The correct option is D $9\sqrt{3}metres$


Let AB be the original height of the tree. Suppose it got bent at a point C and let the part CB take the position CD, meeting the ground at D. Then,



AD = 9 m , $\angle ADC={30}^{\circ }$ and CD = CB

Let AC = x metres and CD = CB = y metres

From right $\Delta DAC$, we have

$\frac{AC}{AD}=tan{30}^{\circ }=\frac{1}{\sqrt{3}}\Rightarrow \frac{x}{9}=\frac{1}{\sqrt{3}}$

$\Rightarrow x=\frac{9}{\sqrt{3}}\Rightarrow x=\frac{9}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=3\sqrt{3}$

Also, from right $\Delta DAC$, we have

$\frac{CD}{AD}=sec{30}^{\circ }=\frac{2}{\sqrt{3}}\Rightarrow \frac{y}{9}=\frac{2}{\sqrt{3}}$

$\Rightarrow y=\frac{18}{\sqrt{3}}\Rightarrow y=\frac{18}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=6\sqrt{3}$

$\therefore AC=3\sqrt{3}m$ and $CB=6\sqrt{3}m$

Total height of the tree $=3\sqrt{3}m+6\sqrt{3}m=9\sqrt{3}metres$