How is the extinction angle for a half wave rectifier with RL load calculated?
For the uncontrolled single-phase rectifier with RL load: In order to find the \$\beta\$, the extinction angle, the following equation must be solved:

$$sin(\beta-\phi)+ sin(\phi)e^{^-\frac{\beta}{\omega\tau}} = 0$$
where  \$tan\phi = \frac{\omega L}{R}\$, Also, can be written as:
$$sin(\beta-\phi)+ sin(\phi)e^{^-\frac{\beta}{tan\phi}} = 0$$

The equation is transcedental and can be solved only by numeric methods.
Using the values you supplied: \$ \phi \approx 37\$ degrees. Solving equation numerically: \$\beta \approx 217\$.
Alternatively, \$\beta(\phi)\$ can be determined by using pre-built plots, like the one shown below (in accordance with the calculation above):



The approximation you mentioned (also in another post similar to this, which you deleted) does not seem always to work . Note, for \$0 \leq \phi\leq 40\$ degrees the curve seems a straight line, being enough to do \$\beta=\phi+180\$ degrees. But, using the plot, for \$\phi=70\$ degrees, \$\beta=260\$. Your approximation lead to \$\beta=250\$, something different.