Question

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

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

Answer 1

Answer: (C)

120${}^o$

In the given figure,

$\angle ACB=\angle ADB={60}^0$ $\left(Angleinsamesegment\right)$

Therefore,

$\angle ADB={60}^0$

We know that the sum of all angles of the triangle is ${180}^0$.

$\angle ADB+\angle DAB+\angle ABD={180}^0$

$\angle DAB+\angle ABD={180}^0-\angle ADB$

$\angle DAB+\angle ABD={180}^0-{60}^0={120}^0$

Hence, this is the answer.