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