Geometry construction

Question

I appreciate any help. I want to find the angle $ADC$. I have drawn the circle in Geogebra, and the angle $ADC=120^\circ.$ But how can I give an argument that is always will be $120^\circ$ if angle $BSC = 60^\circ.$ We see that angle $BAC = 30^\circ$, and angle $ADB = 90^\circ.$ How do I go further? Thank you.

Answer 1

Notice that if you draw the line $CB$, we have that the $\triangle SCB$ is equilateral. Thus, $\angle SBC = 60^\circ$.

Now, the quadrilateral $DABC$ is a cyclic quadrilateral , thus the angle which is opposite to the $\angle SBC$ will be: $$\angle ADC = 180^\circ - 60^\circ = 120^\circ.$$

Answer 2

Your angle $ADC$ is exactly an angle inscribed then its measure is equal to half the measure of the intercepted arc which is clearly equal to $60^{\circ}+180^{\circ}=240^{\circ}$. Consequently your angle is equal to $$120^{\circ}$$