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