Set up the circles so that their centers are on a horizontal line, one to the left and one to the right, and the centers $A$ and $B$ are distance $R$ apart, with $A$ being the left one. There are two intersection points between the circles, an upper, $U$, and a lower, $V$.
Then the distance from $A$ to $U$ is $R$, as is the distance from $B$ to $U$. And the distance from $A$ to $B$ is $R$ as the left circle has radius $R$ and the point $B$ is on it. Hence the triangle $AUB$ is equilateral and has all angles $\pi/3$. The same goes for $AVB$, so the angles at $A$ and at $B$ (namely $VAU$ and $VBU$ must be $2\pi/3$.