# prove that sum of all 3 angles of a triangle is 180.

To prove this we are given a triangle PQR and Angle 1, Angle 2 and Angle 3 are the angles of triangle PQR.

To prove Angle 1 + A angle 2 + Angle 3 = 180 degree, draw a line XYP parallel to QR through the opposite vertex P.

XPY is a line,

Therefore, Angle 4 + Angle 1 + angle 5 = 180 degree                              (1)

XPY || QR and PQ,PR are transvesals

Angle 4 = Angle 2   &

Angle 5 = Angle 3 ( Pairs of alternate angles)

Substituting Anggle 4 and Angle 5 in (1) we get

Angle 2 + Angle 1 + Angle 3 = 180 dgree

i.e. Angle 1 + Angle 2 + Angle 3 = 180 degree

• 6

property ==  [ n - 2] x 180

n=number of sides

in the case of triangle it has 3 sides soo............

=[3-2] x 1800

= 1 x1800 == 1800

• -16

Consider a triangle PQR and ∠1, ∠2 and ∠3 are the angles of ΔPQR (figure shown below). We need to prove that ∠1 + ∠2 + ∠3 = 180°. XPY is a line.
∴∠4 + ∠1 + ∠5 = 180°  … (1)
But XPY || QR and PQ, PR are transversals.
So, ∠4 = ∠2 and ∠5 = ∠3  (Pairs of alternate angles)
Substituting ∠4 and ∠5 in (1), we get
∠2 + ∠1 + ∠3 = 180°
∴∠1 + ∠2 + ∠3 = 180°

• 0

You have done a smal mistake.the queation is like this: Prove that the sum of all angles of a triangle is 180 degree.

Consider a triangle PQR and ∠1, ∠2 and ∠3 are the angles of ΔPQR (figure shown below). We need to prove that ∠1 + ∠2 + ∠3 = 180�. XPY is a line.
∴∠4 + ∠1 + ∠5 = 180� … (1)
But XPY || QR and PQ, PR are transversals.
So, ∠4 = ∠2 and ∠5 = ∠3 (Pairs of alternate angles)
Substituting ∠4 and ∠5 in (1), we get
∠2 + ∠1 + ∠3 = 180�
∴∠1 + ∠2 + ∠3 = 180�
• 88
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