Question
Fri July 06, 2012 By: Priyanshi Shrivastava

to prove that the triangle abc is right angled

Expert Reply
Sat July 07, 2012
We know that for any triangle A+B+C= 180 degrees
cos2A + cos2B +cos2C = 1
C=180 -(A+B)
cos2C = cos2(180-(A+B))=cos2(A+B)= (cosAcosB-sinAsinB)2
=cos2Acos2B + sin2Asin2B-2cosAcosBsinAsinB
=cos2Acos2B + (1-cos2A)(1-cos2B) -2cosAcosBsinAsinB
=cos2Acos2B + 1 - cos2A -cos2B + cos2Acos2B - 2cosAcosBsinAsinB
Now, on substitution
1+2cos2Acos2B-2cosAcosBsinAsinB=1
cosAcosB (cosAcosB-sinAsinB)=0
cos(A+B).cosAcosB=0
Thus either cosA=0 or cosB=0 or cos(A+B)=0
Which reults into making either A,B or A+B = 90 degrees
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