[tex]\displaystyle\bf\\m(\angle A)+m(\angle B)=180^o-m(\angle C)=180^o-40^o=140^0\\\\m(\angle BAI)=\frac{m(\angle A)}{2}\\\\m(\angle ABI)=\frac{m(\angle B)}{2}\\\\m(\angle BAI)+m(\angle ABI)=\frac{m(\angle A)}{2}+\frac{m(\angle B)}{2}=\\\\=\frac{m(\angle A)+m(\angle B)}{2}=\frac{140^o}{2}=70^o\\\\m(\angle AIB)=180-(m(\angle BAI)+m(\angle ABI))=180^o-70^o=110^o\\\\\implies~\boxed{\bf m(\angle AIB)=110^o}[/tex]
#copaceibrainly
