Question

Prove that the sum of exterior angles of a triangle obtained by extending its sides in same direction is 360°​

Answer 1

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Answer 2

Question

Prove that the sum of exterior angles of a triangle obtained by extending its sides in same direction is 360°.

Given

• ∠PAN , ∠AND and ∠ACT are exterior angles of triangle ABC

To prove

∠PAB + ∠QBC + ∠ACR = 360°

Proof

Considering exterior ∠PAB OF triangle ABC,

∠ABC and ∠ACB are the it's remote interior angles

∠PAB = ∠ABC + ∠ACB ----(I)

Similarly,

∠ACR = ∠ABC + ∠BAC ----(II)..theorem of remote interior angles and ∠CBQ = ∠BAC + ∠ACB----(III)

Adding (I) , (II) and (III)

∠PAB + ∠ACR + ∠CBQ

= ∠ABC + ∠ACB + ∠ABC + ∠BAC + ∠ACB

= 2∠ABC + 2∠ACB + 2∠BAC

= 2 ( ∠ ABC + ∠ ACB + ∠BAC )

= 2 × 180°...Sum of interior angles of a triangle

= 360°