Question

Question 123

If the sides of a triangle are produced in an order, show that the sum of the exterior angles so formed is ${360}^{\circ }$

Answer 1

In $\Delta ABC$, by exterior angle property,
Exterior $\angle 1=$ Interior $\angle A$ + Interior $\angle B)$ ....... (i)
Exterior $\angle 2=$ Interior $\angle B$ + Interior $\angle C$ ....... (ii)
Exterior $\angle 3=$ Interior $\angle A$ + Interior $\angle C$ ....... (iii)

On adding Eqs. (i), (ii) and (iii), we get
$\angle 1+\angle 2+\angle 3=2(\angle A+\angle B+\angle C)$
[by angle sum property of a triangle, $\angle A+\angle B+\angle C={180}^{\circ }$]
$\Rightarrow \angle 1+\angle 2+\angle 3=2\times {180}^{\circ }\Rightarrow \angle 1+\angle 2+\angle 3={360}^{\circ }$
Hence, the sum of exterior angles is ${360}^{\circ }$