# Better way to find the angle subtended by OI at the vertex.

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??

• Which vertex? $A$? – Marco Vergamini May 11 at 23:17
• @MarcoVergamini yes. – hyphen May 11 at 23:20
• $BAO=\pi/2-B, BAI=A/2, A+B+C=\pi, OAI=BAI-BAO$. – Marco Vergamini May 11 at 23:23
• in the formula $OI^2 = R^2 -2Rr$ O is circumcentre, not orthocentre – liaombro May 12 at 8:31
• @liaombro yes I edited that. – hyphen May 13 at 17:16

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.