Question

Side BC of a triangle ABC has been produced to a point D such that $\angle ACD={120}^{\circ }.$ If $\angle B=\frac{1}{2}\angle A,$ then $\angle A$is equal to

Answer 1

Side BC of $\Delta ABC$ is produced t D, then

Ext. $\angle ACB=\angle A+\angle B$

(Exterior angle of a triangle is equal to the su of its interior opposite angles)

${120}^{\circ }=\angle A+\angle B$ $(\because \angle ACD={130}^{\circ })$

=$\angle A+\frac{1}{2}\angle A$ $(\because \angle B=\frac{1}{2}\angle A)$

=$\frac{3}{2}\angle A$

$\therefore \angle A={120}^{\circ }\times \frac{2}{3}={80}^{\circ }$ (a)