Do AD là phân giác của góc $\overset{\scriptscriptstyle\frown}{A}={{70}^{0}}$ nên góc $\overset\frown{DAB}=\overset\frown{DAC}={{35}^{0}}$.Có góc $\overset\frown{ABC}={{180}^{\circ }}\text{ }\overset{\scriptscriptstyle\frown}{A}\text{ }\text{ }\overset{\scriptscriptstyle\frown}{C}\text{ }=\text{ }{{180}^{0}}-\text{ }{{70}^{0}}-\text{ }{{30}^{0}}=\text{ }{{80}^{0}}$

 Xét tam giác ABD có:

$\overset\frown{ADB}\text{ }=\text{ 18}{{0}^{0}}\text{ }\overset\frown{DAB}\text{ }\text{ }\overset\frown{ABD}$

 $=\text{ 18}{{\text{0}}^{0}}\text{ }\overset\frown{DAB}\text{ }\text{ }\overset\frown{ABC}$

$={{180}^{0}}-{{35}^{0}}-{{80}^{0}}$

$={{65}^{0}}$

Vậy đáp án đúng là A