Question

Exercitiul 16 va rog. Dau coroana.

Answer 1

    
[tex]\displaystyle \frac{\overbrace{m(AB)}}{2} = \frac{\overbrace{m(BC)}}{3} \\ \\ \overbrace{m(AC)}= \frac{1}{3} \overbrace{m(BC)}~\Rightarrow~ \frac{\overbrace{m(AC)}}{1} = \frac{\overbrace{m(BC)}}{3} \\ \\ \Rightarrow~\frac{\overbrace{m(AB)}}{2} = \frac{\overbrace{m(BC)}}{3}=\frac{\overbrace{m(AC)}}{1}=k \\ \\ \overbrace{m(AB)}=2k \\ \overbrace{m(BC)}=3k \\ \overbrace{m(AC)}=k [/tex]

[tex]\displaystyle \overbrace{m(AB)}+ \overbrace{m(BC)}+ \overbrace{m(AC)}=360^o \\ 2k+3k+k = 360^o \\ 6k=360^o \\ k= \frac{360}{6} = 60 \\ \overbrace{m(AB)} =2k = 2\times 60 = 120^o\\ \overbrace{m(BC)}= 3k=3\times 60 = 180^o\\ \overbrace{m(AC)}=k=60^o \\ \text{Masura unui arc de cerc = unghiul la centru care subantinde arcul.} \\ \text{Unghiurile triunghiului inscris in cerc sunt unghiuri inscrise in cerc. }[/tex]

[tex] \text{Un unghi inscris incerc este egal cu jumatate din unghiul la centru } \\ \text{care subintinde acelasi arc.} \\ Rezulta: \\ \widehat{ACB}= \frac{\overbrace{AB}}{2}= \frac{120}{2}=\boxed{60^o} \\ \widehat{BAC}= \frac{\overbrace{BC}}{2}= \frac{180}{2}=\boxed{90^o} \\ \widehat{ABC}= \frac{\overbrace{AC}}{2}= \frac{60}{2}=\boxed{30^o} \\ [/tex]