$\displaystyle \begin{array}{{>{\displaystyle}l}} Có:\ AH=\sqrt{AB^{2} -BH^{2}} =16\ ( Theo\ Pytago)\\ \ \ \ \ \ \ \ \Rightarrow CH=16\\ \Rightarrow AC=\sqrt{AH^{2} +HC^{2}} =16\sqrt{2} \ ( theo\ Pytago)\\ Ta\ có:\ tan\ \mathrm{\widehat{HAB}} =\frac{12}{16} \Rightarrow \mathrm{\widehat{HAB}} =36,87^{o}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ tan\ \mathrm{\widehat{CAH}} =\frac{16}{16} =1\Rightarrow \mathrm{\widehat{CAH}} =45^{o}\\ Vậy\ \mathrm{\widehat{CAB}} =36,86^{o} +45^{o} =81,87^{o} \end{array}$
