Oblique shock wave

Normal component of the upstream Mach number to the oblique shock wave,

$$M_{n1}=M_{1}Sin\beta =4\left ( Sin30^{\circ} \right )=2$$

From Normal shock wave properties table, for $M_{n1}=2$

$\frac{p_{2}}{p_{1}}=4.5$, $\frac{T_{2}}{T_{1}}=1.687$,  $\frac{p_{02}}{p_{01}}=0.7209$, $M_{n2}=0.5774$

At an altitude of $9000\, m$ , $p_{1}=3.08\times 10^{4} N/m^{2}$, $T_{1}=229.74K$

Therefore,

$$P_{2}=\left ( \frac{P_{2}}{P_{1}} \right )P_{1}=\left ( 4.5 \right )\left ( 3.08\times 10^{4} \right)= 13.86\times 10^{4} N/m^{2}$$

$$T_{2}=\left ( \frac{T_{2}}{T_{1}} \right )T_{1}=\left ( 1.687 \right )229.74=387.571K$$

From the $\theta -\beta -M$ diagram , for, wave angle of $\beta =30^{\circ}$ and Mach number of $4$, deflection angle, $\theta =17.7^{\circ}$.

Therefore,

$$M_{2}=\frac{M_{n2}}{Sin\left ( \beta -\theta \right )}=\frac{0.5774}{Sin\left ( 30^{\circ} -17.7^{\circ}\right )}=2.710$$

From the table of isentropic flow properties , for

$$M_{1}=4,\frac{p_{01}}{p_{1}}=151.8,\frac{T_{01}}{T_{1}}=4.2$$

Total pressure,

$$p_{02}=\left ( \frac{p_{02}}{p_{01}} \right )\left ( \frac{p_{01}}{p_{1}} \right )p_{1}=\left ( 0.0709 \right )\left ( 151.8 \right )\left ( 3.08\times 10^{4} \right )=331488.696N/m^{2}$$

Total temperature,

$$T_{02}=T_{01}=\left ( \frac{T_{01}}{T_{1}} \right )T_{1}=4.2\times 229.74K=964.908K$$