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Question

In a ∆ABC, if cosAa+cosBb+cosCc=k a2+b2+c2, then k = ___________________.

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Solution

Using cosine formula

cosA= b2+c2-a22bc, cosB=a2+c2-b22ac and cosC=a2+b2-c22abL.H.S=cosA a+cosB b+cosCc=b2+c2-a22bc×1a+a2+c2-b22ac×1b+a2+b2-c22ab×1c= b2+c2-a2+a2+c2-b2+a2+b2-c2 2abc= a2+b2+c2 2abc=ka2+b2+c2 = R.H.S given k = 12abc

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