In , = and ray AX bisects the exterior angle . If = 70^ , then =

MCQ

In $\triangle\text{ABC}, \ \angle\text{B} = \angle\text{C}$ and ray $AX$ bisects the exterior angle $\triangle\text{DAC}.$ If $\triangle\text{DAX} = 70^\circ,$ then $\angle\text{ACB} =$

A
$55^\circ$
B
$35^\circ$
✓
$70^\circ$
D
$90^\circ$

✓

Answer

Correct option: C.

$70^\circ$

$AX$ is bisector of $\triangle\text{DAC}$
$\Rightarrow\ \angle\text{DAX}=\angle\text{XAC}=70^\circ$
$\Rightarrow\ \angle\text{DAC}=2\times70=140^\circ$
Now, $\angle\text{A} = 180^\circ - \angle\text{DAC} = 180^\circ\text{c}- 180° = 40^\circ$
In $\triangle\text{ABC}$
$\angle\text{A} + \angle\text{B} + \angle\text{C} = 180^\circ$
$\Rightarrow\ 40^\circ + \angle\text{B} + \angle\text{C}= 180^\circ$
$\Rightarrow\ 40^\circ + \angle\text{C} + \angle\text{C}= 180^\circ...(\angle\text{B} = \angle\text{C})$
$\Rightarrow\ 2\angle\text{C} = 140^\circ$
$\Rightarrow\ \angle\text{C} = 70^\circ$
i.e. $\angle\text{ACB} = 70^\circ$