\(\widehat{BIC}=180^0-\widehat{IBC}-\widehat{ICB}=180^0-\dfrac{1}{2}\widehat{ABC}-\dfrac{1}{2}\widehat{ACB}=180^0-\dfrac{1}{2}\left(\widehat{ABC}+\widehat{ACB}\right)=180^0-\dfrac{1}{2}\left(180^0-\widehat{A}\right)=180^0-\dfrac{1}{2}\left(180^0-40^0\right)=110^0\)
Chọn C