= Solution
A <streamline> satisfies
$$
\frac{dr}{r\,d\theta}=\frac{u_r}{u_\theta}
=-\cot(\gamma\theta).
$$
Integration gives
$$
\log r=-\frac1\gamma\log\sin(\gamma\theta)+C.
$$
The radius is smallest on the wedge bisector $\theta=\alpha/2$, where $\sin(\gamma\theta)=1$. Labelling that radius by $r_{\min}$ therefore gives
$$
\boxed{r(\theta)=r_{\min}
\left(\sin\frac{\pi\theta}{\alpha}\right)^{-\alpha/\pi}}.
$$
Every streamline arrives from infinity along one wall, reaches its closest point on the bisector, and returns to infinity along the other wall:
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ib/paper-1-wedge-streamlines.png]
{title=<Streamlines> in a wedge}
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