Question

show that sum of exterior angles of a triangle is 360°.

Answer 1

, SubjectMatterExpert

Member since Sep 21 2013

Sol:
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

HOLA

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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°  
∠2 + ∠b = 180°  
∠3 + ∠c = 180°  

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°.

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