Question

In the given figure, $\angle$ $CAB=$ ${80}^0$, $\angle$ $ABC=$ ${40}^0$. The sum of $\angle$ $DAB+$ $\angle$ $ABD$ is equal to

• A. ${80}^0$
• B. ${100}^0$
• C. ${120}^0$
• D. ${140}^0$

Answer 1

The correct option is C ${120}^0$

Consider the $△CAB.$

The total sum of any triangle $=180°$

$\therefore \angle ACB+\angle CAB+\angle ABC=180°$

$\angle ACB+80+40=180°$

$\therefore \angle ACB=60°$

$AD$ is the bisector of $\angle A.$

$\therefore \angle DAB=40°$

Similarly BC is the bisector of $\angle ABD$

$\therefore \angle ABD=80°$

$\therefore \angle DAB+\angle ABD=40+80$

$=120°$

Hence, the answer is $120°.$