# Proved that the sum of three angleof triangle is 180digree also find the angles of triangles if they are in ratio 5:6:7

2
by sumit31

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by sumit31

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∴∠A=∠1,∠B=∠2,∠C=∠3

and the angles formed by the construction of the parallel line,

∠SAB=∠4 and ∠PAC=∠5

∴∠A+∠SAB+∠PAC=180° (∵straight angle)

∵∠SAB=∠B and ∠PAC=∠C (∵alternate interior angle)

∴∠A+∠B+∠C=180°

hence proved.

2) let the angles be 5a,6a and 7a

∠A=5a, ∠B=6a and ∠C=7a

∴∠A+∠B+∠C=180° (angle sum property of a Δ)

∴5a+6a+7a=180°

⇒18a=180°

⇒a=180/18

∴ a= 10

∴the angles are,

5a=5*10=50°

6a=6*10=60°

7a=7*10=70°

hope that this will help u.

5x+6x+7x=180degrees. (since sum of 3 angles in a triangle is 180 degrees)

18x=180.

x= 10

therefore angles 5x,6x,7x are 50,60,70