Question Number 212457 by 281981 last updated on 14/Oct/24
Answered by som(math1967) last updated on 14/Oct/24
$$\:\begin{vmatrix}{{a}}&{{b}}&{{c}}\\{{b}}&{{c}}&{{a}}\\{{c}}&{{a}}&{{b}}\end{vmatrix}=\mathrm{0} \\ $$$$\Rightarrow{a}\left({bc}−{a}^{\mathrm{2}} \right)−{b}\left({b}^{\mathrm{2}} −{ca}\right)+{c}\left({ab}−{c}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\Rightarrow{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} −\mathrm{3}{abc}=\mathrm{0} \\ $$$$\Rightarrow\left({a}+{b}+{c}\right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{ab}−{bc}−{ca}\right)=\mathrm{0} \\ $$$$\because\:\left({a}+{b}+{c}\right)\neq\mathrm{0} \\ $$$$\therefore{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{ab}−{bc}−{ca}=\mathrm{0} \\ $$$$\Rightarrow\left({a}−{b}\right)^{\mathrm{2}} +\left({b}−{c}\right)^{\mathrm{2}} +\left({c}−{a}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\left({a}−{b}\right)^{\mathrm{2}} =\mathrm{0}\Rightarrow{a}={b} \\ $$$$\Rightarrow\left({b}−{c}\right)^{\mathrm{2}} =\mathrm{0}\Rightarrow{b}={c} \\ $$$$\therefore{a}={b}={c} \\ $$$$\therefore\:\angle{A}=\angle{B}=\angle{C}=\mathrm{60}° \\ $$$${cosAcosB}+{cosBcosC}+{cosCcosA} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$
Commented by 281981 last updated on 14/Oct/24
$${tnq}\:{sir} \\ $$