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If $\sin A:\cos A=4:7$,then the value of $\large\frac{7\sin A-3\cos A}{7\sin A+2\cos A}$ is

$\begin{array}{1 1}(A)\;\large\frac{3}{14}&(B)\;\large\frac{3}{2}\\(C)\;\large\frac{1}{3}&(D)\;\large\frac{1}{6}\end{array} $

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$\large\frac{7\sin A-3\cos A}{7\sin A+2\cos A}$ is $\large\frac{1}{6}$
Hence (D) is the correct answer.
answered Apr 15, 2014 by sreemathi.v
 

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