Answers

2015-09-25T10:47:17+05:30
1) draw a ΔABC then draw a line parallel to BC so that the alternate interior angles becomes same ,name the the end points of the parallel be Sand P 
                         ∴∠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. 

 
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2015-09-25T10:52:49+05:30
Asume 3 angles as 5x,6x,7x
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
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