Question

Calculati raza cercului circumscris triunghiului ABC cu m(<A) =45° si m(<B) =30° iar AB=4.

Answer 1

$\it m(\hat{C}) = 180^o-(45^o+30^o) = 105^o \\ \\ \\ Th. \ sinusurilor \ \Rightarrow \dfrac{c}{sinC}=2R \Rightarrow \dfrac{4}{sin105^o} = 2R|_{:2} \Rightarrow \\ \\ \\ \Rightarrow R=\dfrac{2}{sin105^o} \qquad (1) \\ \\ \\ sin105^o = sin(180^o-105^o) =sin75^o = sin(45^o+30^o) = \\ \\ \\ = sin45^ocos30^o+sin30^ocos45^o = \dfrac{\sqrt2}{2}\cdot \dfrac{\sqrt3}{2} + \dfrac{1}{2}\cdot \dfrac{\sqrt2}{2} = \\ \\ \\ =\dfrac{\sqrt6+\sqrt2}{4} \qquad (2)$

$\it (1),\ (2) \Rightarrow R=\dfrac{2}{\dfrac{\sqrt6+\sqrt2}{4}} =2\cdot \dfrac{4} {\sqrt6+\sqrt2} =\dfrac{ 2\cdot4\cdot(\sqrt6-\sqrt2)}{4} = \\ \\ \\ = 2(\sqrt6-\sqrt2)$