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For triangle ABC prove that, cos (B+C/2)= sin A/2 |
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Answer» ngle ABC A°+B°+C°=180° A+B+C/2=180/2 A+B/2=90-C/2 MULTIPLY BY Sin Both Side Sin(A+B/2)=Sin(90-C/2) We Know That Sin(90-A)=COSA So,Sin(A+B/2)=CosC/2 Thanks me MARK me as BRAINLIEST :-) Rate my answers ;) #BrainlyPrince727 |
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