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Question

In triangle ABC, a(b2+c2)cosA+b(c2+a2)cosB+c(a2+b2)cosC=

A
2a2b2c2
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B
3a2b2c2
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C
3abc
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D
2abc
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Solution

The correct option is C 3abc
a(b2+c2)cosA+b(c2+a2)cosB+c(a2+b2)cosC
=ab2cosA+ac2cosA+bc2cosB+ba2cosB+ca2cosC+cb2cosC
=ab(bcosA+acosB)+bc(bcosC+ccosB)+ac(ccosA+acosC)
=abc+abc+abc=3abc [Using projection formula]

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