A new spaceport is proposed for Camden County, Georgia (USA). The latitude is 30.56N and the only available launch azimuth is 123 degrees. What is the resulting LEO without mid-course maneuvers? Can show me how you calculate this?
$$ \cos(inclination) = \cos(lat)*\sin(azimuth) $$
So in your example, the inclination would be equal to: $$ \arccos(\cos(30.56)*\sin(123)) = 43.77\deg $$
To simplify the calculation, I assumed that your given azimuth was an inertial one. The calculations that take into account the rotation of the Earth and what the compass onboard would measure as the azimuth are in the link above.
To amplify on the correct answer of @M.A.H.:
To apply this general spherical triangle (https://en.wikipedia.org/wiki/Solution_of_triangles) to the particular problem of orbital inclination:
Point $C$ is the geographical North Pole; Point $B$ is the launch point, so the arc $C-B$ is the co-latitude of the launch point; $A$ is the point where the orbit track crosses the equator, so the arc $C-A$ is $90^o$ and the orbital inclination is $\alpha$. Once the launch azimuth is related to the angle $\beta$ (depending on the exact geometry) the triangle can be solved for $\alpha$, and also $\gamma$, the longitude of the ascending node of the orbit.
Orbital inclination will be equal to the latitude. The reason is simple - you just need that orbit plane will pass through the center of the Earth and your launch point
See explanation here: Of inclinations and azimuths