Question

A straight tree breaks due to wind and the broken parts bends without getting detached such that the top of the tree touches the ground making an angle of 30${}^o$ with the ground. The distance from the part of the tree to the point when the top touches the ground is 10 m. The height of the tree is _____.

• A. 10($\sqrt{3}$ - 1) m
• B. 10($\sqrt{3}$ + 1) m
• C. 10($\sqrt{3}4$ - 1) m
• D. 10($\sqrt{3}$ m

Answer 1

Answer: (D)

10($\sqrt{3}$ m

Refer figure, let AB be the tree broken at point C

such that CD takes position CD and $\angle$ADC = 30${}^o$.

Then, $AD=10$ m.
Let, $AC=x$ m and CD $=y$ m
Then the height of the tree $=(x+y)$ m

Now, $\frac{AC}{AD}=\tan {30}^o\Rightarrow \frac{x}{10}\Rightarrow x=\frac{10}{\sqrt{3}}$ m

And, $\frac{AD}{CD}=\cos {30}^o\Rightarrow \frac{\sqrt{3}}{2}=\frac{10}{y}\Rightarrow y=\frac{20}{\sqrt{3}}$ m

$\therefore$ Height of tree $=x+y$
= $\frac{10}{\sqrt{3}}+\frac{20}{\sqrt{3}}$ m

= $\frac{30}{\sqrt{3}}$m = 10$\sqrt{3}$m

So, option D is correct.