prove that " in the quadrilateral , the sum of all exterior angle is 360
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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°.
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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°.
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@@ @@ @@ @@ @@ @@
Please mark the answer as brain list answer
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