Question

If sin A : cos A = 4 : 7, then the value of $\frac{7sinA-3cosA}{1sinA+2cosA}$ is ______.

• A. $\frac{7}{18}$
• B. $\frac{3}{2}$
• C. $\frac{1}{3}$
• D. $\frac{1}{6}$

Answer 1

Answer: (A)

$\frac{7}{18}$

$\frac{\sin A}{\cos A}=\frac{4}{7}$, So

$\frac{7\sin A-3\cos A}{1\sin A+2\cos A}=\frac{\frac{7\sin A}{\cos A}-3}{\frac{1\sin A}{\cos A}+2}$

$=\frac{7\times \frac{4}{7}-3}{1\times \frac{4}{7}+2}$

$=\frac{7}{18}$