| If in any triangle

*ABC*, the base*BC*is produced in both ways, the sum of the exterior angles at*B*and*C*is :A. π-A

B. π+A

C. (π/2)+A

D. π–

### Right Answer is: B

#### SOLUTION

*ABD* = *π* – *B*, *ACE* = π–*C* *ABD*+*ACE* = 2*π* -(*B+C*) = 2π – (π – *A*)

= π + *A*