From the top of a tree of height \[120\,\,m,\] the angle of depression of two boats in the same line with the foot of the tree and on the same side of it are \[45{}^\circ \] and \[60{}^\circ ,\]respectively. The distance between the boats is [SSC (10+2) 2013] |
A) \[40\,\,(3-\sqrt{3})\,\,m\]
B) \[40\,\,(3\sqrt{3}-1)\,\,m\]
C) \[120\,\,(\sqrt{3}-1)\,\,m\]
D) \[12\,\,(3-\sqrt{3})\,\,m\]
Correct Answer: A
Solution :
Given, height of the tree \[=120\,\,m\] |
In the given figure two depression angles \[45{}^\circ \]and \[60{}^\circ \]are given, |
Now, \[\angle CAO=\angle ACB=60{}^\circ \][alternate angles] |
Similarly, \[\angle DAO=\angle ADB\] |
In \[\Delta ADB,\]\[\tan 45{}^\circ =\frac{AB}{BD}\]\[\Rightarrow \]\[1=\frac{120}{BD}\] |
\[\Rightarrow \] \[BD=120\,\,m\] |
In \[\Delta ACB,\]\[\tan 60{}^\circ =\frac{AB}{BC}\]\[\Rightarrow \]\[\sqrt{3}=\frac{120}{BC}\] |
\[\Rightarrow \]\[BC=\frac{120}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=40\sqrt{3}\,\,m\] |
Thus, distance between both the boats, |
\[CD=BD-BC\] |
\[=120-40\sqrt{3}=40\,\,(3-\sqrt{3})m\] |
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