A) $ 100{}^\circ $
B) $ 110{}^\circ $
C) $ 120{}^\circ $
D) $ 130{}^\circ $
Correct Answer: D
$ \Rightarrow $ $ 80{}^\circ +\angle B+\angle C=180{}^\circ $
$ \Rightarrow $ $ \angle B+\angle C=180{}^\circ -80{}^\circ $
$ \angle B+\angle C=100{}^\circ $
Divide by 2 on both the sides, we get
$ \frac{\angle B}{2}+\frac{\angle C}{2}=50{}^\circ $
i.e. $ \angle DBC+\angle DCB=50{}^\circ $
$ [ \because \frac{\angle B}{2}=\angle DBCand\frac{\angle C}{2}=\angle DCB ] $
Now, in $ \Delta BCO, $
$ \angle BDC+\angle DBC+\angle DCB=180{}^\circ $
[by angle sum property form Eq. (i)]
$ \Rightarrow $ $ \angle BDC+50{}^\circ =180{}^\circ $
$ \Rightarrow $ $ \angle BDC=180{}^\circ -50{}^\circ =130{}^\circ $ Alternate Method In $ \Delta ABC, $ and BD and CD are internal bisectors of $ \angle B $ and $ \angle C. $
$ \therefore $ $ \angle BDC=90{}^\circ +\frac{1}{2}\angle A $ [by results] $ =90{}^\circ +\frac{1}{2}\times 80{}^\circ =130{}^\circ $
BETA
