If a,b,c are the sides of the ΔABC and a2,b2,c2 are the roots of x3−px2+qx−k=0, then
A
cosAa+cosBb+cosCc=P2√k
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B
acosA+bcosB+ccosC=4q−p22√k
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C
asinA+bsinB+csinC=2pΔ√k
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D
sinAsinBsinC=8Δ3k
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Solution
The correct option is DsinAsinBsinC=8Δ3k (a)cosAa+cosBb+cosCc=b2+c2−a22abc+c2+a2−b22abc+a2+b2−c22abca2+b2+c22abc=P2√k(b)acosA+bcosB+ccosC=a2(b2+c2−a2)+b2(c2+a2−b2)+c2(a2+b2−c2)2abc=4(a2b2+b2c2+c2a2)−(a2+b2+c2)22abc=4q−p22√k (c)asinA+bsinB+csinC=a2+b2+c22R=(a2+b2+c2)4Δ2abc=2pΔ√k.(d)sinAsinBsinC=abc8R3=abc×64Δ38(abc)3=8Δ3k.