Wiemy, że:

$\left|AB\right|=c,\left|BC\right|=a,\left|AC\right|=b$

oraz

$\left|\sphericalangle CAB\right|={45}^{\circ },\left|\sphericalangle ABC\right|={30}^{\circ }$   

zatem:

$\left|\sphericalangle ACB\right|={180}^{\circ }-\left({45}^{\circ }+{30}^{\circ }\right)={180}^{\circ }-{75}^{\circ }={105}^{\circ }$

Wiemy również, że:

$\cos {105}^2=\frac{\sqrt{2}-\sqrt{6}}{4}$

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Wobec tego korzystając z jedynki trygonometrycznej dostajemy: 

${\sin }^2{105}^2+{\cos }^2{105}^{\circ }=1$

${\sin }^2{105}^{\circ }+{\left(\frac{\sqrt{2}-\sqrt{6}}{4}\right)}^2=1$

${\sin }^2{105}^{\circ }+\frac{2-2\sqrt{12}+6}{16}=1$

${\sin }^2{105}^{\circ }+\frac{8-4\sqrt{3}}{16}=1$      

${\sin }^2{105}^{\circ }=1-\frac{4\left(2-\sqrt{3}\right)}{16}$

${\sin }^2{105}^{\circ }=1-\frac{2-\sqrt{3}}{4}$

${\sin }^2{105}^{\circ }=\frac{2+\sqrt{3}}{4}\mid \sqrt{},\sin {105}^{\circ }>0$

$\sin {105}^{\circ }=\frac{\sqrt{2+\sqrt{3}}}{2}$     

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Korzystając z twierdzenia sinusów dostajemy:

1)

$\frac{c}{\sin {105}^{\circ }}=2R$

$\frac{c}{\frac{\sqrt{2+\sqrt{3}}}{2}}=2R$

$c=2R\cdot \frac{\sqrt{2+\sqrt{3}}}{2}=\underline{\underline{\sqrt{2+\sqrt{3}}\cdot R}}$   

2)

$\frac{a}{\sin {45}^{\circ }}=2R$

$\frac{a}{\frac{\sqrt{2}}{2}}=2R$

$a=2R\cdot \frac{\sqrt{2}}{2}=\underline{\underline{\sqrt{2}\cdot R}}$    

3)  

$\frac{b}{\sin {30}^{\circ }}=2R$

$\frac{b}{\frac{1}{2}}=2R$

$b=2R\cdot \frac{1}{2}=\underline{\underline{R}}$