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1) prove that the sum of all angles of a triangle is 180. in an isosceles right triangle , find all the angle of triangle.

2) prove that the sum of the angles of a triangle is two right angles. if in a right triangle an acute angle is one-fourth the other acute angle


Wkt, in isosceles right triangle ,one is 90 degree angle and another two are same...means let it be x so, the eqn. will be .............
so,one angle is of 90 2nd is of 45 and 3rd is also of 45
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yes , but i also needed how to prove that in a triangle , sum of all angles equals to 180
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Draw ABC as any triangle.

Now, draw a straight line parallel to BC through A. 
     Now angle DAG = angle BAC - opposite angles at an intersection of two lines.

     Angle DAE = angle ABC  as they are angles formed by a line AB intersecting
       two parallel lines  - included angle.
     Angle GAF = angle EAC - opposite angles at intersection of two lines at A
     Angle GAF = angle BAC    as BC and FE are parallel. AC is intersecting both
             lines. These are included angles.

     Sum of these angles at A =  angles DAG + GAF + EAD = 180 degrees.
      At any point the total angle 360° degrees is divided into 180° and 180° by a
      straight line.

So sum of angles A+B+C = 180° degrees

ISOSCELES right angle triangle:  let angle B = 90° deg
         90° + angle C + angleA = 180°
         angle C = angle A        isosceles triangle.
           2 angle C = 90°        angle C = angle A = 45°
A+B+C = 180°
B = 90°
A = 1/4 C
So      C = 4 A
Hence A+90°+4A = 180°
5 A = 180-90 = 90

A = 18°
So C = 72°

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