Xét ΔABC có \(\widehat{BAC}+\widehat{ABC}+\widehat{ACB}=180^0\)
=>\(\widehat{ABC}+\widehat{ACB}=180^0-65^0=115^0\)
=>\(2\left(\widehat{IBC}+\widehat{ICB}\right)=115^0\)
=>\(\widehat{IBC}+\widehat{ICB}=\dfrac{115^0}{2}=57,5^0\)
Xét ΔIBC có \(\widehat{IBC}+\widehat{ICB}+\widehat{BIC}=180^0\)
=>\(\widehat{BIC}+57,5^0=180^0\)
=>\(\widehat{BIC}=180^0-57,5^0=122,5^0\)