Question
$ABC$ is a triangle in which $\angle \text{B} = \angle \text{C}$ and ray $AX$ bisects the exterior angle $DAC$. If $\angle \text{DAX}=70^\circ.$ Find $\angle \text{ACB}$
✓
Answer
Here, $\angle \text{CAX}=\angle \text{DAX}$
$($AX bisects $\angle \text{CAD})$
$\angle \text{CAX}=70^\circ$
$\angle \text{CAX}+\angle \text{DAX}+\angle \text{CAB}=180^\circ$
$70^\circ+70^\circ+\angle \text{CAB}=180^\circ$
$\angle \text{CAB}=180^\circ-140^\circ$
$\angle \text{CAB}=40^\circ$
$\angle \text{ACB}+\angle \text{CBA}+\angle \text{CAB}=180^\circ$
$($Sum of the angles of $\triangle \text{ABC})$
$\angle \text{ACB}+\angle \text{ACB}+40^\circ=180^\circ$
$(\angle \text{C}=\angle \text{B})$
$2\angle \text{AVB}=180^\circ-40^\circ$
$\angle \text{ACB}=\frac{140}{2}$
$\angle \text{ACB}=70^\circ$
$($AX bisects $\angle \text{CAD})$
$\angle \text{CAX}=70^\circ$
$\angle \text{CAX}+\angle \text{DAX}+\angle \text{CAB}=180^\circ$
$70^\circ+70^\circ+\angle \text{CAB}=180^\circ$
$\angle \text{CAB}=180^\circ-140^\circ$
$\angle \text{CAB}=40^\circ$
$\angle \text{ACB}+\angle \text{CBA}+\angle \text{CAB}=180^\circ$
$($Sum of the angles of $\triangle \text{ABC})$
$\angle \text{ACB}+\angle \text{ACB}+40^\circ=180^\circ$
$(\angle \text{C}=\angle \text{B})$
$2\angle \text{AVB}=180^\circ-40^\circ$
$\angle \text{ACB}=\frac{140}{2}$
$\angle \text{ACB}=70^\circ$
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