sin(BCA) = [tex] \frac{ \sqrt{3} }{2} [/tex] ==> m(BCA) = 60°
m(BCA) = 60° ===> m(BAC) = 30°
In triunghiul ABC , m(B) = 90° si m(A) = 30° ==>
==> AC = 2BC = 2 · 5 = 10 cm
cos(C) = [tex] \frac{ca}{i} = \frac{BC}{AC} = \frac{5}{10} = \frac{1}{2} [/tex]
In triunghiul ABC aplicam Teorema lui Pitagora
AC² = AB² + BC²
10² = AB² + 5²
AB² = 100 - 25
AB² = 75
AB = 5√3 cm
tg(C) = [tex] \frac{co}{ca} = \frac{AB}{BC} = \frac{5 \sqrt{3} }{5} = \sqrt{3} [/tex]
ctg(C) = [tex] \frac{ca}{co} = \frac{BC}{AB} = \frac{5}{5 \sqrt{3} } = \frac{1}{ \sqrt{3}} [/tex]
