Question

In a $\Delta ABC$, the sum of exterior angles at $B$ and $C$ is equal to:

• A. (a)${180}^{\circ }-\angle BAC$
• B. (b)${180}^{\circ }+\angle BAC$
• C. (c)${180}^{\circ }-2\angle BAC$
• D. (d)${180}^{\circ }+2\angle BAC$

Answer 1

Answer: (B)

(b)${180}^{\circ }+\angle BAC$

Theorem: An measure of an exterior angle of a triangle is equal to the sum of the measures of the two non-adjacent interior angles.
So In our case $\angle CAB$ and $\angle CBA$ are the two non-adjacent interior angles corresponding to the exterior angle.$\angle DCA$. and
$\angle ACB$ and $\angle BAC$ are the two non-adjacent interior angles corresponding to the exterior angle.$\angle ABE$.
Hence, $\angle CAB$$+$$\angle CBA$$=$$\angle DCA$..........(1)
And,
$\angle CAB$$+$$\angle BCA$$=$$\angle ABE$..........(2)
Adding (1) and (2) we get,
$\angle ABE+\angle DCA={180}^{\circ }+\angle BAC$