Solution
= Solution
The required transmitted <P wave> angle would satisfy
$$
\sin i_{P2}=\frac{3500}{1500}\sin30^\circ=\frac76>1.
$$
Therefore \b[there is no real transmitted P-ray angle]. The incidence exceeds the P critical angle $\arcsin(1500/3500)=25.38^\circ$; the transmitted P contribution is <evanescent> away from the interface rather than a propagating body ray. Reporting $90^\circ$ would describe critical incidence, not the stated $30^\circ$ incidence.