Question

Aflati cos B in triunghiul ABC unde AC= 4 si raza cercului circumscris triunghiului este R=12.

Answer 1

 

[tex]\displaystyle\\ \frac{AC}{\sin B} = 2R~~~~~~~\text{(Teorema sinusurilor)}\\\\ \sin B=\frac{AC}{2R}=\frac{4}{2\times 12}=\frac{4}{24}=\boxed{\frac{1}{6}}\\\\ \cos^2B=1-\sin^2B = 1-\left(\frac{1}{6}\right)^2 = 1 - \frac{1}{36}=\frac{35}{36}\\\\ \cos B = \sqrt{\frac{35}{36}}\\\\ \boxed{\bf \cos B = \frac{\sqrt{35}}{6}}[/tex]