The circle in the diagram has centre \(C\). Three angles \(\alpha, \beta, \gamma\) are also indicated.

Circle with centre C. Alpha is the angle at a vertex A between a line segment tangent to the circle and a line segment ending at C. This line segment AC is also part of a triangle whose other vertex, B, lies on the circle and has angle gamma. The angle of the triangle at vertex A, adjacent to alpha, is beta.

The angles \(\alpha,\beta, \gamma\) are related by the equation:

  1. \(\cos \alpha = \sin (\beta + \gamma)\),

  2. \(\sin \beta = \sin \alpha \sin \gamma\),

  3. \(\sin \beta (1 − \cos \alpha) = \sin \gamma\),

  4. \(\sin (\alpha + \beta) = \cos \gamma \sin \alpha\).

Let’s give the vertices in the diagram some names. We’ll call the vertex of angles \(\alpha\) and \(\beta\), \(A\), and the vertex of angle \(\gamma\), B.

We’ll also draw in the radius between C and the tangent - they’ll be perpendicular at the point of contact. We’ll call this point \(D\).

The same diagram as above but with the additional points A,B, and D marked

By looking at the right-angled triangle \(ACD\), we can see that the length of \(AC = x = \dfrac{r}{\sin \alpha}\), so \(r = x\sin\alpha\).

Now we can apply the sine rule to the triangle \(ABC\) to give \[\frac{\sin \beta}{r}=\frac{\sin \gamma}{x},\] so \[x\sin\beta = r\sin\gamma \implies x\sin\beta = x\sin\alpha\sin\gamma.\]

So on cancelling \(x\) we have \[\sin \beta = \sin \alpha \sin \gamma,\] and the answer is (b).