Solution:
In quad. \[ABCD,\]
\[\angle DAB\text{ }+\angle DCB\text{ }=\text{ }{{180}^{o}}\]
\[{{27}^{o}}~+\angle CAB\text{ }+\text{ }{{83}^{o}}~=\text{ }{{180}^{o}}\]
Hence,
\[\angle CAB\text{ }=\text{ }{{180}^{o}}-\text{ }{{110}^{o}}~=\text{ }{{70}^{o}}\]