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Question Number 212457 by 281981 last updated on 14/Oct/24

Answered by som(math1967) last updated on 14/Oct/24

$$\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|=0$$$$\Rightarrow a(bc-a^2)-b(b^2-ca)+c(ab-c^2)=0$$$$\Rightarrow a^3+b^3+c^3-3abc=0$$$$\Rightarrow (a+b+c)(a^2+b^2+c^2-ab-bc-ca)=0$$$$\because (a+b+c)\ne 0$$$$\therefore a^2+b^2+c^2-ab-bc-ca=0$$$$\Rightarrow {(a-b)}^2+{(b-c)}^2+{(c-a)}^2=0$$$$\Rightarrow {(a-b)}^2=0\Rightarrow a=b$$$$\Rightarrow {(b-c)}^2=0\Rightarrow b=c$$$$\therefore a=b=c$$$$\therefore \mathrm{\angle }A=\mathrm{\angle }B=\mathrm{\angle }C=60°$$$$cosAcosB+cosBcosC+cosCcosA$$$$=\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{3}{4}$$

Commented by 281981 last updated on 14/Oct/24

$$tnqsir$$

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