# If cosA+cosC=sinB then prove that ABC is right angled triangle

Samar Zaidi
15 Points
5 years ago
You mentioned that ABC is right angle triangle.Lets do some Hit and TrialB=90A=30C=60Then Cos A+ Cos C is not Equal to Sin BWhen we mention that ABC is a right angle triangle then B=90Hence we should mention that either ACB is a right angle triangle or CAB is right angle triangle. Then we can prove it
Harshit Sengar
27 Points
5 years ago
See as ABC is a triangle then, A+B+C=180° similarly, A+C=180°-B and B+C=180°-A Then, CosA+CosC=SinB2 Cos(A+C/2) * Cos(A-C/2)=2SinB/2 * CosB/22Cos(180°-B/2) * Cos(A-C/2)=2SinB/2 * CosB/2Cos(A-C/2)=CosB/2A-C=BA=B+CA=180°-A2A=180° Therefore, A=90°Hence proved
Samar Zaidi
15 Points
5 years ago
I am not doubting the answer, i also got it but what i mentioned is that if you say that ABC is a right angle triangle then by default B=90
Harshit Sengar
27 Points
5 years ago
See man if any angle in a triangle is 90° then the triangle is also right angled triangle. The triangle ABC can also be determined as: B / | / | / | /____ | C A
Jashan
13 Points
5 years ago
CosA+cosC=sinBCosA+cosC=cos(90-B)So A+C=B We know that A+B+C=180Now put A+C=BTherefore B+B=1802B=180 B=90
Soumendu Majumdar
159 Points
5 years ago
cos A + cos C = sin B
Now using the formula cos X + cos Y = 2 cos(X + Y)/2 cos(X – Y)/2
so L.H.S= 2{ cos(A + C)/2 }{ cos(A – C)/2 }
A + B + C = 180 degrees
so B/2 = 90 – (A + C)/2
so L.H.S= 2sin(B/2){ cos(A – C)/2 }
Using sin 2X = 2 sin X cos X
R.H.S = 2 sin (B/2) cos (B/2)
So from L.H.S & R.H.S
we get
cos(A – C)/2 = cos B/2
so A – C = B
implies 2 A = A + B + C = 180 degrees
so A = 90 degrees
Hence BAC is a right angled triangle