Determine the reading of the middle hours on December 3 at the time of Arcturus (and Bootes) at an altitude of 45o in the eastern

Determine the reading of the middle hours on December 3 at the time of Arcturus (and Bootes) at an altitude of 45o in the eastern part of the sky in St. Petersburg. And also find the azimuth of Arcturus at this moment

To determine the average hour reading, you need to know the sidereal hour reading. And in order to know the sidereal clock, you need to determine the hour angle of the star from the conditions of the problem and the position of the star in the sky.
From the table we have: the latitude of Petersburg is φ = + 60 °, and for Arcturus a = 14h, b = + 20 °.
Having constructed a celestial sphere for the city and drawing an almucantarat 45о and a celestial parallel 20о, we will determine the position of Arcturus in the sky by the intersection of these circles, i.e., point M.
Now let’s draw a circle of declination RMP, find the hour angle of Arcturus at this moment: t = NEQR = 22h (approximately). Then sidereal time at this moment will be:
s = t + a = 22h + 14h = 36h or 12h
By calculation, we find that the sidereal time on November 3 at the middle noon will be 17h, and we got the sidereal time equal to 12h. Therefore, the observed position of Arcturus corresponded to the moment 5 o’clock before noon (17h – 12h = 5h), i.e. 7 o’clock in the morning. Thus, the desired reading of the average hours is 7 o’clock in the morning.
Drawing the vertical ZMK, we will determine the azimuth of Arcturus at this moment; it will be: A = NSK = 60o east (approx).

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