Răspuns:
Explicație pas cu pas:
a) Aplicăm Teorema sinusului
[tex]\dfrac{a}{sinA}= \dfrac{b}{sinB}=\dfrac{c}{sinC}=2R,~=>~\dfrac{a+b+c}{sinA+sinB+sinC}=2R,~=>\\=>~\dfrac{P}{sinA+sinB+sinC}=2R,~=>~ sinA+sinB+sinC=\dfrac{P}{2R}=\dfrac{2p}{2R}=\dfrac{p}{R}[/tex]
unde p este semiperimetrul triunghiului, iar R raza cercului circumscris.
[tex]b)~ctg\frac{x}{2}=-\frac{1}{2},~dar~tg\frac{x}{2}=\dfrac{1}{ctg\frac{x}{2} }=\dfrac{1}{-\frac{1}{2} } =-2.\\sinx=\dfrac{2tg\frac{x}{2}}{1+tg^2\frac{x}{2}}=\dfrac{2*(-2)}{1+(-2)^2} =-\dfrac{4}{5}\\cosx=\dfrac{1-tg^2\frac{x}{2}}{1+tg^2\frac{x}{2}}=\dfrac{1-(-2)^2}{1+(-2)^2}=\dfrac{1-4}{1+4}=-\dfrac{3}{5}.\\F=2*(-\dfrac{3}{5})-3*(-\dfrac{4}{5})+7=-\dfrac{6}{5}+\dfrac{12}{5}+7=\dfrac{6}{5}+7=1,2+7=8,2.[/tex]
