question_answer

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,

\[CD=BD-BC\]

\[=120-40\sqrt{3}=40\,\,(3-\sqrt{3})m\]