5. ecuatia dreptei AB este (x-xA)/(xB-xA)=(y-yA)/(yB-yA), adica (x-2)/(1-2)=(y-5)/(3-5), rezulta (x-2)/(-1)=(y-5)/(-2) => -2x+4= -y+5, deci y=2x+1. Punctul C apartine dreptei AB => coordonatele lui C, adica m si 1 verifica ecuatia dreptei AB: y=2x+1, adica 1=2m+1, deci m=0
6.E([tex]\frac{ \pi }{3} [/tex])= cos([tex] \frac{ \pi }{6} [/tex])+sin([tex] \frac{ \pi }{3} [/tex]) = [tex] \frac{ \sqrt{3}}{2} [/tex] + [tex] \frac{ \sqrt{3}}{2} [/tex] = [tex] \sqrt{3} [/tex]
