Question

If $\tan \alpha =\frac{4}{15}(0<\alpha <{90}^{\circ })$, then the value of $\frac{5\sin \alpha +7\cos \alpha }{6\cos \alpha -3\sin \alpha }$

• A. $\frac{125}{78}$
• B. $\frac{30}{3}$
• C. $\frac{125}{102}$
• D. None of these

Answer 1

The correct option is A $\frac{125}{78}$

Given, $\tan \alpha =\frac{4}{15}$
Thus we know $\sin \alpha =\frac{4}{\sqrt{4^2+{15}^2}}$
and $\cos \alpha =\frac{15}{\sqrt{4^2+{15}^2}}$
Therefore, $\frac{5\sin \alpha +7\cos \alpha }{6\cos \alpha -3\sin \alpha }=\frac{5\times 4+7\times 15}{6\times 15-3\times 4}=\frac{125}{78}$