If \(a\) is the variable of the horizontal axis, then you can solve for \(b\) to get the equation of the line in slope-intercept form in the \(a,b\) plane:
\(-a+3b=0\implies b=\dfrac a3\)
i.e. a line with slope \(\dfrac13\) through the origin, which means it is contained in the first and third quadrants. Since the terminal side of \(\theta\) has a negative sine, the angle must lie in the third quadrant.
Because the slope of the line is \(\dfrac13\), you can choose any length along the line to make up the hypotenuse of a right triangle with reference angle \(\theta\). Any such right triangle will have \(\tan\theta=\dfrac13\), regardless of whether the angle is the first or third quadrant. But since \(\theta\) is known to lie in the third quadrant, and so \(\sin\theta\) and \(\cos\theta\) are both negative, you have
\(\tan\theta=\dfrac{\sin\theta}{\cos\theta}=\dfrac13\implies \dfrac{\sin\theta}{\cos\theta}=\dfrac{-1}{-3}\implies \cos\theta=-3\)
