I presume that I can remove the capacitors since the circuit uses DC, but then I don't understand why I can apply a voltage divider in the 30K and 20K resistors since the base is connected between them.
I don't understand why I can apply a voltage divider in the 30K and 20K resistors since the base is connected between them.
The emitter resistor of 1 kohm projects an impedance onto the base that is beta times higher. So, if beta is 100 and the emitter resistor is 1 kohm, the base will look like a loading impedance of 100 kohm. This loading resistance is seen in parallel with the 20 kohm resistor thus, the effective resistance at the base to ground is 20 kohm || 100 kohm = 16.67 kohm. You can use this to determine a more precise value of base voltage to ground.
I estimate 12 volts * 16.67/(30 + 16.67) = 4.29 volts.
What your prof does is find the equivalent series resistance of the 20 kohm and 30 kohm resistors and to do that you parallel them (12 kohm = 20 k || 30 k). He then calculates the open circuit voltage (base disconnected) as being 12 * 20/(20 + 30) = 4.8 volts.