Question

In a $\Delta ABC,\angle A={50}^{\circ }$ and BC is produced to a point D. If the bisectors of $\angle ABC$ and $\angle ACD$ meet at E, then $\angle E$=

Answer 1

In $\Delta ABC,\angle A={50}^{\circ }$

BC is produced

Bisectors of $\angle ABC$ and $\angle ACD$ meet at $\angle E$

$\therefore \angle E=\frac{1}{2}\angle A=\frac{1}{2}\times {50}^{\circ }={25}^{\circ }$