A) $ \frac{1}{4} $
B) $ \frac{1}{6} $
C) $ \frac{1}{3} $
D) $ \frac{2}{3} $
Correct Answer: B
$ \Rightarrow $ $ 5\sin \theta =4cos\theta $ … (i) Now, $ \frac{5\sin \theta -3\cos \theta }{5\sin \theta +2\cos \theta }=\frac{4\cos \theta -3\cos \theta }{4\cos \theta +2\cos \theta } $ [from Eq. (i)] $ =\frac{\cos \theta }{6\cos \theta }=\frac{1}{6} $ Alternate Method Given, $ 5\tan \theta =4 $
$ \Rightarrow $ $ \tan \theta =\frac{4}{5} $ $ \because $ $ \tan \theta =\frac{BC}{AB} $ Similarly,
$ \therefore $ $ \sin \theta =\frac{4}{\sqrt{41}} $
$ \Rightarrow $ $ \cos \theta =\frac{5}{\sqrt{41}} $
$ \therefore $ $ \frac{5\sin \theta -3\cos \theta }{5\sin \theta +2\cos \theta }=\frac{5\times \frac{4}{\sqrt{41}}-3\times \frac{5}{\sqrt{41}}}{5\times \frac{4}{\sqrt{41}}+2\times \frac{5}{\sqrt{41}}} $
[putting values]
$ =\frac{20-15}{20+10}=\frac{5}{30}=\frac{1}{6} $
BETA
