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Dear student,
COS3B+COS3C+COS3A=1 COS3B+COS3C+COS3A-1=0 COS3(B+C)/2.COS3(B-C)/2-SIN^3A/2=0 SOLVING WE GET COS3C/2.COS3B/2.COS3A/2=0 EITHER3C/2= C=/3 THEREFORE TRIANGLE IS ACUTE ANGLED
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