Question

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

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

Answer 1

The correct option is C ${120}^{\circ }$

In $△ACB$
Sum of angles = 180
$\angle ACB+\angle ABC+\angle BAC=180$
$\angle ACB=180-80-40$
$\angle ACB={60}^{\circ }$
$\angle ACB=\angle ADB={60}^{\circ }$ (Angles in a same segment)
Now, in $△ADB$
sum of angles = 180
$\angle ADB+\angle ABD+\angle BAD=180$
$60+\angle ABD+\angle BAD=180$
$\angle ABD+\angle BAD={120}^{\circ }$