Question

If the sides of a triangle are produced in order, then the sum of the three exterior angles so formed is:

• A. ${90}^{\circ }$
• B. ${180}^{\circ }$
• C. ${270}^{\circ }$
• D. ${360}^{\circ }$

Answer 1

The correct option is D

${360}^{\circ }$

In $\Delta ABC,$ sides AB, BC and CA are produced in order, then exterior $\angle FAB,\angle DBC$ and $\angle ACE$ are formed

We know an exterior angles of a triangle is equal to the sum of its interior opposite angles

$\therefore \angle FAB=\angle B+\angle C$

$\angle DBC=\angle C+\angle A$ and

$\angle ACE=\angle A+\angle B$

Adding we get,

$\angle FAB+\angle DBC+\angle ACE=\angle B+\angle C+\angle C+\angle A+\angle A+\angle B$

=2$(\angle A+\angle B+\angle C)$

=2$\times {180}^{\circ }$ (Sum of angles of a triangle)

=${360}^{\circ }$