\(\widehat{A}=180^0-120^0=60^0\\ \Rightarrow2\cdot60^0=120^0=3\widehat{B}\\ \Rightarrow\widehat{B}=40^0\\ \Rightarrow\widehat{BIA}=180^0-\widehat{IBA}-\widehat{IAB}=180^0-\dfrac{1}{2}\widehat{A}-\dfrac{1}{2}\widehat{B}=180^0-\dfrac{1}{2}\left(\widehat{A}+\widehat{B}\right)=180^0-\dfrac{1}{2}\cdot100^0=130^0\)