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If $\theta$ lies in the first quadrant and $5\;\tan \theta=4$, then $\large\frac{5 \sin \theta -3 \cos \theta}{\sin \theta+ 2 \cos \theta}$ is equal to

\[\begin {array} {1 1} (a)\;\frac{5}{14} & \quad (b)\;\frac{3}{14} \\ (c)\;\frac{1}{14} & \quad (d)\;0 \end {array}\]

1 Answer

$(a) \frac{5}{14}$
answered Nov 7, 2013 by pady_1
 

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