Question

Prove that the sum of the exterior angles of a triangle is 360°

Answer 1

Consider ΔABC in which ∠A = 1, ∠B = 2 and ∠C = 3 Let the exterior angles of A, B and C be ∠a, ∠b and ∠c respectively. Recall that sum of angles in a triangle is 180° That is  ∠1 + ∠2 + ∠3 = 180° From the figure, we have ∠1 + ∠a = 180°  [Linear pair] ∠2 + ∠b = 180°  [Linear pair] ∠3 + ∠c = 180°  [Linear pair] Add the above three equations, we get ∠1 + ∠a + ∠2 + ∠b + ∠3 + ∠c = 180° + 180° + 180°  ⇒ (∠1 + ∠2 + ∠3) + ∠a + ∠b + ∠c = 540° ⇒ 180°+ ∠a + ∠b + ∠c = 540° ⇒ ∠a + ∠b + ∠c = 540° – 180° = 360° Thus sum of exterior angles of a triangle is 360°.

Answer 2

Question-

Prove that the sum of the exterior angles of a triangle is equal to 360 degrees.

Answer-

Let us say, ∠1, ∠2 and ∠3 are the interior angles of a triangle. When we extend the sides of the triangle in the outward direction, then the three exterior angles formed are ∠4, ∠5 and ∠6, which are consecutive to ∠1, ∠2 and ∠3, respectively.

Hence,

• ∠1 + ∠4 = 180°   ……(i)
• ∠2 + ∠5 = 180°  …..(ii)
• ∠3 + ∠6 = 180°  …..(iii)

If we add the above three equations, we get;

∠1+∠2+∠3+∠4+∠5+∠6 = 180° + 180° + 180°

Now, by angle sum property we know,  

∠1+∠2+∠3 = 180°

Therefore,  

180 + ∠4+∠5+∠6 = 180° + 180° + 180°

⇒ ∠4+∠5+∠6 = 360°

This proves that the sum of the exterior angles of a triangle is equal to 360 degrees.