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 \[BD+DC\] |
\[=\frac{15}{\sqrt{3}}+\frac{30}{\sqrt{3}}=\frac{45}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=15\sqrt{3}\,\,m\] |
\[=15\times 17.3=25.95\,\,m\] |
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