Question

$\angle CAB={80}^0,\angle ABC={40}^0$ then $\angle DAB+\angle ABD$ is

• A. 80
• B. 100
• C. 120
• D. 140

Answer 1

The correct option is C 120

According to the problem,

$\angle CAB+\angle CBA+\angle ACB={180}^{\circ }$

${80}^{\circ }+{40}^{\circ }+\angle ACB={180}^{\circ }$

$\angle ACB=180-120$

$\angle ACB=60$

$\angle ACB=\angle ADB$

Now,

$\angle ADB+\angle DAB+\angle ABD={180}^{\circ }$

$60+\angle DAB+\angle ABD=180$

$\angle DAB+\angle ABD=180-60$

$\angle DAB+\angle ABD={120}^{\circ }$