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Show that the line joining the incenter to the circumcentre of triangle ABC is inclined to the side BC at an angle tan^(-1) ((cos B + cos C -1)/(sin C - sin B)) |
Answer» Solution :Let `I` be the incenter and O be the circumcenter of the triangle ABC. Let OL be PARALLEL to BC Let `angle IOL = theta` Now, `IM = r, OC = R, angle NOC = A`. Then `tan theta = (IL)/(OL) = (IM- LM)/(BM- BN)` `= (IM - ON)/(BM - NC)` `= (r - R cos A)/(r cot. (B)/(2) - R SIN A)` `=(4R sin.(A)/(2) sin.(B)/(2) sin.(C)/(2) - R COSA)/(4R sin.(A)/(2) sin.(B)/(2) sin.(C)/(2). cot. (B)/(2)- R sin A)` `=(cos A + cos B + cos C - 1- cos A)/(sin A + sin C - sin B - sin A)` `= (cos B + cos C -1)/(sin C - sin B)` or `theta = tan^(-1) [(cos B + cos C -1)/(sin C - sin B)]` |
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