Question

In $\Delta ABC,BD$ bisects angle B. If $\angle A=\frac{2}{3}\angle B$ and $\angle B=3\angle C$, then $\angle BDC$ is:

• A. ${75}^o$
• B. ${105}^o$
• C. ${90}^o$
• D. ${120}^o$

Answer 1

The correct option is B ${105}^o$

$\angle A=\frac{2}{3}\angle Bor\angle B=\frac{3}{2}\angle A$
Again, $\angle A=\frac{2}{3}\angle B=\frac{2}{3}\times 3\angle Cor\angle C=\frac{\angle A}{2}$
$\angle A+\angle B+\angle C={180}^o$
Substituting the values of $\angle Band\angle C$
$\angle A+\frac{3}{2}\angle A+\frac{\angle A}{2}={180}^o$
$or\angle A={60}^o$
$\therefore \angle B=\frac{3}{2}\angle A=\frac{3}{2}\times {60}^o={90}^o$
$\angle BDC=\angle A+\frac{1}{2}\angle B={60}^o+\frac{1}{2}\times {90}^o={105}^o$