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 $\angle\text{DAC}.$ If $\angle\text{DAX}=70^\circ$ then $\angle\text{ACB}=$

A
35º
B
90º
C
70º
D
55º

✓

Answer

70º
Solution:

AX is bisector of $\angle\text{DAC}.$
$\Rightarrow\angle\text{DAX}=\angle\text{XAC}=70^\circ$
$\Rightarrow\angle\text{DAC}=2\times70^\circ=140^\circ$
Now, $\angle\text{A}=180^\circ-\angle\text{DAC}=180^\circ-140^\circ=40^\circ$
In $\triangle\text{ABC},$
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow40^\circ+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow40^\circ+\angle\text{C}+\angle\text{C}=180^\circ\dots(\angle\text{B}=\angle\text{C})$
$\Rightarrow2\angle\text{C}=140^\circ$
$\Rightarrow\angle\text{C}=70^\circ$
$\text{i.e}\angle\text{ACB}=70^\circ$

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