prove that the sum of all exterior angles of a triangle is 360
Answers
Answer:
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°.
hope it helps you
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°