question_answer

If 5 tan A = 4, then the value of \[\frac{5\sin A-3\cos A}{\sin A+2\,\cos A}\] is                                 [SSC (10+2) 2015]

A) \[\frac{9}{14}\]                         

B) \[\frac{5}{6}\]

C) \[\frac{7}{9}\]                          

D) \[\frac{5}{14}\]

Correct Answer: D

Solution :

\[5\tan A=4\]\[\Rightarrow \]\[\tan A=\frac{4}{5}\]

\[\therefore \]      \[AC=\sqrt{16+25}=\sqrt{41}\]

Now, \[\frac{5\sin A-3\cos A}{\sin A+2\cos A}\]

\[=\frac{5\times \frac{4}{\sqrt{41}}-3\times \frac{5}{\sqrt{41}}}{\frac{4}{\sqrt{41}}+2\times \frac{5}{\sqrt{41}}}=\frac{20-15}{4+10}=\frac{5}{14}\]