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Question

Let a,b and c be the sides of a ABC. If a2,b2,c2 are the roots of the equation x3Px2+QxR=0, where P,Q,R are constants, then find the value of cosAA+cosBB+cosCC in terms of P,Q and R.

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Solution

Given, cosAa+cosBb+cosCc---------------------------(1)
Applying cosine rule:
cosA=b2+c2a22bc,cosB=a2+c2b22ac,cosC=a2+b2c22ab
putting values in equation (1)
=b2+c2a22abc+a2+c2b22abc+a2+b2c22abc
=(a2+b2+c2)2abc
Now,
X3Px2+QxR=0
Sum of roots, a2+b2+c2=P
Product of roots, a2b2c2=R
Then,(a2+b2+c2)2abc = p2R
cosAa+cosBb+cosCc=P2R


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