question_answer

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\[\sqrt{3}=1.73\])

A) 21.45 m           

B) 25.95 m

C) 27.25 m           

D) 28.15 m

Correct Answer: B

Solution :

Let AB be the tree bent at point D, so that DA takes the position DC.

Then,    \[DA=DC\]

\[BC=15\,\,m\]

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 \[BD+DC\]

\[=\frac{15}{\sqrt{3}}+\frac{30}{\sqrt{3}}=\frac{45}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=15\sqrt{3}\,\,m\]

\[=15\times 17.3=25.95\,\,m\]