Proof that sum of three angles are 180 degree
Answers
Answered by
1
draw a parallel line to BC passing through the point A.
As Q is line the SUM of angles PAB,QAC,BAC is 180°.
PAB=ABC
QAC=ACB
BAC=BAC
PAB+QAC+BAC=180°
SUSTITUTE THE GIVEN VALUES
ABC+ACB+BAC=180°
HENCE PROVED
As Q is line the SUM of angles PAB,QAC,BAC is 180°.
PAB=ABC
QAC=ACB
BAC=BAC
PAB+QAC+BAC=180°
SUSTITUTE THE GIVEN VALUES
ABC+ACB+BAC=180°
HENCE PROVED
Attachments:
Answered by
3
Draw line a through points A and B. Draw line b through point C and parallel to line a.
Since lines a and b are parallel, <)BAC = <)B'CA and <)ABC = <)BCA'.
It is obvious that <)B'CA + <)ACB + <)BCA' = 180 degrees.
Thus <)ABC + <)BCA + <)CAB = 180 degrees.
Similar questions