Question

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

• A. ${35}^{\circ }$
• B. ${90}^{\circ }$
• C. ${70}^{\circ }$
• D. ${55}^{\circ }$

Answer 1

The correct option is C

${70}^{\circ }$

In $\Delta ABC,\angle B=\angle C$

AX is the bisector of ext. $\angle CAD$

$\angle DAX={70}^{\circ }$

$\therefore \angle DAC={70}^{\circ }\times 2={140}^{\circ }$

But Ext. $\angle DAC=\angle B+\angle C$

=$\angle C+\angle C$ $(\therefore \angle B=\angle C)$

=$2\angle C$

$\therefore 2\angle C={140}^{\circ }\Rightarrow \angle C=\frac{{140}^{\circ }}{2}={70}^{\circ }$

$\therefore \angle ACB={70}^{\circ }$