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Question

In triangle â–³ABC, with usual notation, then
cosA=b2+c2−a22bc, cosB=c2+a2−b22ca,cosC=a2+b2−c22ab

A
True
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B
False
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Solution

The correct option is A True
According to the cosine rule in any triangle

a2=b2+c22bccosA
So
CosA=b2+c2a22bc

Similarly CosB=c2+a2b22ca

and CosC=a2+b2c22ab

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