The pressure jump relation with $p_2/p_1=1+\epsilon$ gives
$$
M_{x1}^2=1+\frac{\gamma+1}{2\gamma}\epsilon.
$$
If $\beta_1$ is the angle between the upstream velocity and the shock front, then $M_{x1}=M_1\sin\beta_1$. To leading order,
$$
\boxed{\beta_1\simeq\beta_2\simeq
\arcsin(M_1^{-1})},
$$
the <Mach angle>. Expansion of the compression ratio gives
$$
\frac{\rho_2}{\rho_1}=1+\frac\epsilon\gamma+O(\epsilon^2).
$$
Because $u_y$ is continuous and $\tan\beta=u_x/u_y$,
$$
\boxed{\frac{\tan\beta_1}{\tan\beta_2}
=\frac{u_{x1}}{u_{x2}}
=1+\frac\epsilon\gamma+O(\epsilon^2)}.
$$
Writing $\theta=\beta_1-\beta_2$ and linearizing the tangent about $\sin\beta_1=1/M_1$ gives the <weak-oblique-shock deflection>
$$
\boxed{\theta\simeq
\frac\epsilon\gamma\sin\beta_1\cos\beta_1
=\frac{\epsilon\sqrt{M_1^2-1}}{\gamma M_1^2}}.
$$
The normal component decreases while the tangential component is unchanged, so $\beta_2<\beta_1$: the flow turns toward the shock front.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 314 / 2 / d / Solution

If $p_2/p_1=1+\epsilon$ with $0<\epsilon\ll1$ and the upstream Mach number is $M_1>1$, the shock angle is the Mach angle to leading order and the flow deflection is
$$
\theta\simeq\frac{\epsilon}{\gamma}
\frac{\sqrt{M_1^2-1}}{M_1^2}.
$$
The normal speed decreases, so the downstream flow turns toward the shock front.

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 Incoming links: Weak-oblique-shock deflection