Since the line segment PQ is tantgent to the circle, then the angle mTherefore we know that the angle OQR has to be 30°. This is shown in the figure:
Now, we know that the sum of all the internal angles of a triangle must be 180°, then for the triangle RQP we have:
![30+\beta+x=180](https://img.qammunity.org/2023/formulas/mathematics/college/ovdmz2f7zxrg80dme7chzi8l9x30g96r9w.png)
For the triangle OQR we have
![\alpha+\gamma+60=180](https://img.qammunity.org/2023/formulas/mathematics/college/zljo8uoqgvtjmrqzj9hq8iidfjgjzrea3a.png)
and for the triangle OQP we have
![x+\alpha+90=180](https://img.qammunity.org/2023/formulas/mathematics/college/c5gelpl8o9lqdd4m7a3mm87wvvehh7x9mc.png)
Summing up we have the system
![\begin{gathered} x+\beta+30=180 \\ \alpha+\gamma+60=180 \\ x+\alpha+90=180 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/fuvggudte18p5drooe7z7mi3zmin1r5u7k.png)
At first glance we see that this system hava more unknown variables than equations, but we have to notice that the angles alpha and gamma are supplementary, then
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