Question Number 106019 by ZiYangLee last updated on 02/Aug/20
Answered by ajfour last updated on 02/Aug/20
$$\frac{\mathrm{3}}{{a}+{b}+{c}}−\frac{{a}+\left({a}+{b}+{c}\right)}{\left({a}+{b}\right)\left({a}+{c}\right)}=\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{3}\left({a}+{b}\right)\left({a}+{c}\right)−{a}\left({a}+{b}+{c}\right) \\ $$$$\:\:\:\:\:\:−\left({a}+{b}+{c}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$${a}^{\mathrm{2}} +{bc}−{b}^{\mathrm{2}} −{c}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\:\:\mathrm{cos}\:{A}=\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{bc}}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\:\:{A}=\mathrm{60}°\:. \\ $$$${So}\:{reverse}\:{is}\:{true}\:{as}\:{well}. \\ $$
Commented by ZiYangLee last updated on 02/Aug/20
$$ \\ $$$$\mathrm{WoW}\:\mathrm{THaNkS} \\ $$