Better way to find the angle subtended by OI at the vertex A.
O is circumcentere and I is incenter of a triangle ABC.
So I did solved this problem by finding the distance OI which is $ \sqrt{ R^2 - 2Rr} $ and then I used cosine rule to find angle subtended . Which is $ \frac{B-C}{2} $ but that took a lot of work. Is there any other better way??
Its either $\frac{A}{2} + B - 90$ or $\frac{A}{2} + C - 90$ depending on the orientation of the bisector of A and the line from A normal to BC.