$\triangle{PQR}$ is inscribed in circle C such that the measure of $\angle{PRQ}$'s intercepted arc is $70^o$ and m$\angle{PQR} = 50^o$. Find the measure of $\angle{QPR}$'s intercepted arc.
When I read the problem, the triangle that I constructed was inscribed in C with all of the vertices touched the circle. However, I wasn't sure what the difference between $\angle{PRQ}$ and $\angle{PRQ}$'s intercepted arc. It seems to me that they are the same thing, but apparently not.
An intercepted arc is the central length that subtends the intercepted angle. The relationship between the intercepted arc and intercepted angle is $(intercepted \space arc) = 2 \times (intercepted \space angle)$. The angle $\angle{PRQ}$'s intercepted arc then yields a intercepted angle of $35^o = \angle{PRQ}$. Since $\angle{PRQ}$ and $\angle{PQR}$ are known for the triangle, we can then find the last side, $\angle{QPR} = 95^o$. This intercepted arc for this angle is then $2\times{95} = 190^o$
The solution to this problem confirms my result. Everyone who responded to give idea, thanks :)