Avem ΔABC dreptunghic in A (<A = 90°)
sin <B = b/a = 2/3
cos <B = c/a
Aplicam teorema lui Pitagora:
b² + c² = a²
[tex] \frac{b^{2}+c^{2}}{a^{2}}=1<=> \frac{b^{2}}{a^{2}}+\frac{c^{2}}{a^{2}}=1 \\ \frac{c^{2}}{a^{2}}=1-\frac{b^{2}}{a^{2}}=> \frac{c}{a}= \sqrt{1-\frac{b^{2}}{a^{2}}}=\sqrt{1-\frac{2^{2}}{3^{2}}} = \\ = \sqrt{1-\frac{4}{9}}= \sqrt{ \frac{5}{9}}= \frac{ \sqrt{5} }{3}[/tex]
=> cos <B = √5 / 3
