Quantitative Aptitude Ques 1207
Question: The upper part of a tree is broken by wind into two parts makes an angle of $ 30{}^\circ $ with the ground. The top of the tree touches the ground at a distance of 15 m from the foot of the tree. Find the height of the tree before it was broken. (Take $ \sqrt{3}=1.73 $ )
Options:
A) 21.45 m
B) 25.95 m
C) 27.25 m
D) 28.15 m
Show Answer
Answer:
Correct Answer: B
Solution:
- Let AB be the tree bent at point D, so that DA takes the position DC. Then, $ DA=DC $ $ BC=15m $ and $ \angle BCD=30{}^\circ $ In right angled $ \Delta CBD, $
$ \Rightarrow $ $ \tan 30{}^\circ =\frac{BD}{BC} $
$ \Rightarrow $ $ \frac{1}{\sqrt{3}}=\frac{BD}{15} $
$ \Rightarrow $ $ BD=\frac{15}{\sqrt{3}}m $
In right angled $ \Delta CBD, $
$ \Rightarrow $ $ \cos 30{}^\circ =\frac{BC}{CD} $
$ \Rightarrow $ $ \frac{\sqrt{3}}{2}=\frac{15}{CD} $
$ \Rightarrow $ $ CD=\frac{30}{\sqrt{3}}m $
$ \therefore $ Total length of the tree $ BD+DC $ $ =\frac{15}{\sqrt{3}}+\frac{30}{\sqrt{3}}=\frac{45}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=15\sqrt{3}m $ $ =15\times 17.3=25.95m $
BETA
