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Let $U=|u_1|$ and $q=(\gamma -1)/(\gamma +1)$, the strong-shock limit of the normal velocity ratio. In the plane containing the upstream direction and the shock normal, take $u_1=U{e}_X$ and $n=\cos \beta e_X+\sin \beta e_Y$. The normal velocity is multiplied by $q$ while the tangential velocity is unchanged, henceEliminating $\beta$ gives the strong-shock polar for a perfect gas:Equivalently this is the circleBoth signs of the perpendicular component are possible by changing shock orientation. The endpoints $u_{X2}=qU,U$ lie on the upstream axis. An original sketch for $\gamma =5/3$ illustrates the locus and a tangent ray:

The deflection is the angle of the ray from the origin to the downstream velocity, rather than the shock-normal angle $\beta$. Its maximum occurs where this ray is tangent to the circle. The circle's center is at distance $\gamma U/(\gamma +1)$ and its radius is $U/(\gamma +1)$, so the tangent right triangle provesThis is the maximum deflection through a strong perfect-gas shock; for $\gamma =5/3$ it is about $36.87$ degrees.