Quantitative Aptitude Ques 1775
Question: If $ 2\cot \theta =3, $ then what is $ \frac{2\cos \theta -\sin \theta }{2\cos \theta +\sin \theta } $ equal to?
Options:
A) $ \frac{2}{3} $
B) $ \frac{1}{3} $
C) $ \frac{1}{2} $
D) $ \frac{3}{4} $
Show Answer
Answer:
Correct Answer: C
Solution:
- In $ \Delta ABC, $ $ \cot =\frac{3}{2}=\frac{AB}{AC} $
$ \therefore $ $ AB=3 $ and $ AC=2 $
By Pythagoras theorem,
$ BC^{2}={{(2)}^{2}}+{{(3)}^{2}} $
$ \Rightarrow $ $ BC=\sqrt{13} $ Now, $ \cos \theta =\frac{Base}{Hypotenuse}=\frac{AB}{AC}=\frac{3}{\sqrt{13}} $ and $ \sin \theta =\frac{Perpendicular}{Hypotenuse}=\frac{AC}{BC}=\frac{2}{\sqrt{13}} $
$ \therefore $ $ \frac{2\cos \theta -\sin \theta }{2\cos \theta +\sin \theta }=\frac{\frac{6}{\sqrt{13}}+\frac{2}{\sqrt{13}}}{\frac{6}{\sqrt{13}}+\frac{2}{\sqrt{13}}}=\frac{4}{8}=\frac{1}{2} $
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