$\phi = 40^\circ$ N, $\delta = 20^\circ$ N, $H = 30^\circ$ (west). $\sin a = \sin40\sin20 + \cos40\cos20\cos30 \approx 0.2198 + 0.6634 = 0.8832$ → $a \approx 62.0^\circ$. $\cos A = (\sin20 - \sin40\sin62)/(\cos40\cos62) \approx (0.3420 - 0.588)/0.359 \approx -0.685$, $\sin A = (\sin30 \cos20)/\cos62 \approx (0.5*0.9397)/0.4695 \approx 1.001$ → $A \approx 135^\circ$ (southeast? Wait, sin positive, cos negative → quadrant II → $A \approx 180-43=137^\circ$). So azimuth ≈ 137° from north.
How do we find a star's current local position based on its universal coordinates, the observer's latitude, and the time? The Solution: spherical triangle spherical astronomy problems and solutions
From the cosine formula, setting $h=0$: $$ 0 = \sin \phi \sin \delta + \cos \phi \cos \delta \cos H $$ $$ \cos H = - \frac\sin \phi \sin \delta\cos \phi \cos \delta $$ Or simplified: $$ \cos H = - \tan \phi \tan \delta $$ $\phi = 40^\circ$ N, $\delta = 20^\circ$ N,