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 $\angle \text{ACB}$
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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 $\triangle \text{ABC})$
$\angle \text{ACB}+\angle \text{ACB}+40^\circ=180^\circ$ $(\angle \text{C}=\angle \text{B})$
$2\angle \text{AVB}=180^\circ-40^\circ$
$\angle \text{ACB}=\frac{140}{2}$
$\angle \text{ACB}=70^\circ$
$\angle \text{CAX}=\angle \text{DAX}$ $($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 $\triangle \text{ABC})$
$\angle \text{ACB}+\angle \text{ACB}+40^\circ=180^\circ$ $(\angle \text{C}=\angle \text{B})$
$2\angle \text{AVB}=180^\circ-40^\circ$
$\angle \text{ACB}=\frac{140}{2}$
$\angle \text{ACB}=70^\circ$
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