Răspuns:
p.s. aplicate proportii derivate...
Explicație pas cu pas:
Prin AC, AB, BC vom înțelege arce ...
(AC, AB, BC) i.p. (0,(3), 0,1(6), 0,(1)), ⇒
[tex]\dfrac{AC}{0,(3)}= \dfrac{AB}{0,1(6)}=\dfrac{AB}{0,(1)}, ~~deoarece~0,(3)=\frac{3}{9}=\frac{1}{3},~~0,1(6)=\frac{16-1}{90}=\frac{15}{90}=\frac{1}{6},~~0,(1)=\frac{1}{9},~obtinem \\\dfrac{AC}{\frac{1}{3}}=\dfrac{AB}{\frac{1}{6}}=\dfrac{BC}{\frac{1}{9}}=\dfrac{AC+AB+BC}{\frac{1}{3}+\frac{1}{6}+\frac{1}{9}}=\dfrac{360}{\frac{6+3+2}{18} }=\dfrac{360*18}{11} \\Deci,~~AC=\dfrac{1}{3}*\dfrac{360*18}{11}=\dfrac{360*6}{11}=\dfrac{2160}{11}\\[/tex]
[tex]AB=\dfrac{1}{6}*\dfrac{360*18}{11}=\dfrac{360*3}{11}=\dfrac{1080}{11}\\BC=\dfrac{1}{9}*\dfrac{360*18}{11}=\dfrac{360*2}{11}=\dfrac{720}{11}[/tex]
