In this particular question, there is a special case where the half line is a tangent when $\theta=\pi$, so the answer can be found using triangles. How would the type of question be solved when there is not a trivial solution?, e.g. if in this case, the half line was $arg(z-3+2i)=\theta$
Given solution:
For an arbitrary point $P(a,b)$, we have $AO = (-6-a)+i(-6-b)$, which makes an angle $$\theta=\tan^{-1}\frac{6+b}{6+a}\tag 1$$ with the positive $x$ axis. Also, we have
$$\sin\frac{\alpha}2 = \frac{|OT|}{|AO|}=\frac4{\sqrt{(6+a^2)^2+(6+b)^2}}$$
or,
$$\frac{\alpha}2 = \sin^{-1}\frac4{\sqrt{(6+a^2)^2+(6+b)^2}}\tag 2$$
Then, there will be no common solution for $\arg(z-(a+ib))$ to be outside the range
$$ \theta + \frac{\alpha}2 < \arg(z-(a+ib)) < \theta - \frac{\alpha}2 $$
where $\theta$ and $\alpha$ are given by (1) and (2) respectively.