Question Number 15894 by Tinkutara last updated on 15/Jun/17
$$\mathrm{If}\:\mathrm{the}\:\mathrm{angles}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangle}\:{ABC}\:\mathrm{be}\:\mathrm{in} \\ $$$$\mathrm{A}.\mathrm{P}.,\:\mathrm{then} \\ $$$$\left(\mathrm{1}\right)\:{c}^{\mathrm{2}} \:=\:{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \:−\:{ab} \\ $$$$\left(\mathrm{2}\right)\:{b}^{\mathrm{2}} \:=\:{a}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} \:−\:{ac} \\ $$$$\left(\mathrm{3}\right)\:{a}^{\mathrm{2}} \:=\:{b}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} \:−\:{ac} \\ $$$$\left(\mathrm{4}\right)\:{b}^{\mathrm{2}} \:=\:{a}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} \\ $$
Commented by ajfour last updated on 15/Jun/17
$$\left(\mathrm{2}\right)\:{is}\:{correct}\:,\:{if}\:\angle{B}=\pi/\mathrm{3} \\ $$$$\:\:\mathrm{2}{ca}\mathrm{cos}\:{B}={c}^{\mathrm{2}} +{a}^{\mathrm{2}} −{b}^{\mathrm{2}} \\ $$$$\:{with}\:\mathrm{cos}\:{B}=\pi/\mathrm{3} \\ $$$$\:{b}^{\mathrm{2}} ={c}^{\mathrm{2}} +{a}^{\mathrm{2}} −{ac}\:. \\ $$$${And}\:{one}\:{of}\:{the}\:{angles}\:{need}\:{be}\:\pi/\mathrm{3}. \\ $$$${if}\:\mathrm{2}{B}={A}+{C}\:\:{then} \\ $$$$\:\:\:\:\:\mathrm{2}{B}=\pi−{B} \\ $$$$\:\:\:\:\:\:\:\:{B}=\pi/\mathrm{3}\:. \\ $$$${Any}\:{one}\:{of}\:\left(\mathrm{1}\right),\:\left(\mathrm{2}\right),\:{or}\:\left(\mathrm{3}\right)\:{can} \\ $$$${be}\:{correct}\:{depending}\:{on}\:{which} \\ $$$${angle}\:{is}\:\pi/\mathrm{3}\:. \\ $$
Commented by Tinkutara last updated on 16/Jun/17
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$