[tex]a+4b=0\implies a=-4b[/tex]
Treat [tex]a[/tex] as a function of [tex]b[/tex], so that any point on this line takes the form [tex](b,-4b)[/tex]. Suppose [tex]b[/tex] is positive; then any such point lies in the 4th quadrant, and this guarantees that the angle [tex]\theta[/tex] has a positive value for [tex]\cos\theta[/tex].
By definition of tangent and cotangent, we have
[tex]\tan\theta=\dfrac{-4b}b=-4\implies\cot\theta=-\dfrac14[/tex]
Recall the Pythagorean identity,
[tex]\cot^2\theta+1=\csc^2\theta[/tex]
In the 4th quadrant, we have [tex]\sin\theta<0[/tex], so that [tex]\csc\theta<0[/tex] as well. So when we solve for [tex]\csc\theta[/tex] above, we need to take the negative square root:
[tex]\csc\theta=-\sqrt{\cot^2\theta+1}=-\dfrac{\sqrt{17}}4[/tex]
[tex]\implies\sin\theta=-\dfrac4{\sqrt{17}}[/tex]
