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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??

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

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