Question

If $\angle A={60}^o,\angle E={40}^o$ and $\angle C={30}^o,$ then $\angle BDC=$

• A. ${50}^o$
• B. ${100}^o$
• C. ${70}^o$
• D. ${80}^o$

Answer 1

The correct option is A ${50}^o$

We know that, Sum of Opposite angles of the Exterior Angle of a Triangle is equal to the measure of that exterior angle

Here, We can see that $\angle DBC$ is Exterior to $△AEB$

Hence, $\angle A+\angle B=\angle DBC$ (Exterior angle)

$\Rightarrow {60}^o+{40}^o=\angle DBC\Rightarrow \angle DBC={100}^o$

Now, in $△DBC$

$\angle BDC+\angle DBC+\angle C={180}^o$ (Sum of interior angles is ${180}^o$)

$\Rightarrow \angle BDC+{100}^o+{30}^o={180}^o\Rightarrow \angle BDC={50}^o$