Question

Given: $5\cos A-12\sin A=0$,

$\frac{\sin A+\cos A}{2\cos A-\sin A}$ is $\frac{m}{19}$, m is

Answer 1

$5\cos A-12\sin A=0$
$5\cos A=12\sin A$
$\tan A=\frac{5}{12}$
Now, $\frac{\sin A+\cos A}{2\cos A-\sin A}$
Divide by $\cos A$
= $\frac{\tan A+1}{2-\tan A}$
= $\frac{\frac{5}{12}+1}{2-\frac{5}{12}}$
= $\frac{5+12}{24-5}$
= $\frac{17}{19}$