Question

prove that in a quadrilateral the sum of all exterior angles is 360 degree

Answer 1

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Here is your answer;

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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] ________(1)

∠2 + ∠b = 180°  [Linear pair] _______(2)

∠3 + ∠c = 180°  [Linear pair] _______(3)

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 this helps you deaR ✌✌✌.

Answer 2

To prove: ∠P + ∠Q + ∠R + ∠S = 360º

Proof:

Consider triangle PQS, we have,

⇒ ∠P + ∠PQS + ∠PSQ = 180º ... (1)  [Using Angle sum property of Triangle]

Similarly, in triangle QRS, we have,

⇒ ∠SQR + ∠R + ∠QSR = 180º ... (2)  [Using Angle sum property of Triangle]

On adding (1) and (2), we get

∠P + ∠PQS + ∠PSQ + ∠SQR + ∠R + ∠QSR = 180º + 180º

⇒ ∠P + ∠PQS + ∠SQR + ∠R + ∠QSR + ∠PSQ  = 360º

⇒ ∠P + ∠Q + ∠R + ∠S  = 360º  [Hence proved]