Prove that the sum of the angles of a triangle is 180°(not talking about equilateral triangle).
If u know the answer then plz answer it...it is urgent.
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Hello,
Consider a triangle PQR and ∠1, ∠2 and ∠3 are the angles of ΔPQR (see figure).
We need to prove that
∠1 + ∠2 + ∠3 = 180°
We use the properties related to parallel lines to prove this. For this, let us draw a line XPY parallel to QR through the opposite vertex P, as shown in the figure.
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°
then
∠1 + ∠2 + ∠3 = 180°
bye :-)
Consider a triangle PQR and ∠1, ∠2 and ∠3 are the angles of ΔPQR (see figure).
We need to prove that
∠1 + ∠2 + ∠3 = 180°
We use the properties related to parallel lines to prove this. For this, let us draw a line XPY parallel to QR through the opposite vertex P, as shown in the figure.
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°
then
∠1 + ∠2 + ∠3 = 180°
bye :-)
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Dia095:
Thnx
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