if A B and C are interior angles of triangle A B C then show that
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In triangle
A+B+C= 180. By pvt the
C+B=180-A
Now, divid it by 2
C/2+B/2 = 90 - A/2
Put sin on both side
Sin(C/2+B/2 ) = sin( 90- A/2)
Sin(C/2+B/2)=cosA/2. ----- -------reason-CosA= sin(90-A)
Hope this understand
A+B+C= 180. By pvt the
C+B=180-A
Now, divid it by 2
C/2+B/2 = 90 - A/2
Put sin on both side
Sin(C/2+B/2 ) = sin( 90- A/2)
Sin(C/2+B/2)=cosA/2. ----- -------reason-CosA= sin(90-A)
Hope this understand
krantigaynar:
If this help plez mark as brainlist answer!!!!:)
that was very useful thank u
hmm
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