Korzystamy z twierdzenia sinusów:
[tex]sin\alpha = \frac{a}{2R}[/tex]
[tex]a = \sqrt{2}+\sqrt{3}\\R \ - \ promien \ okregu \ opisanego \ na \ dowolnym \ trojkacie[/tex]
[tex]R = \frac{abc}{4P}[/tex]
a,b,c - długości boków trójkata
P - pole trójkąta
[tex]\sqrt{2}^{2}+2^{2} = b^{2}\\\\2+4 = b^{2}\\\\b^{2} = 6\\\\\underline{b = \sqrt{6}}\\\\\sqrt{3}^{2}+2^{2} = c^{2}\\\\3+4 = c^{2}\\\\c^{2} = 7\\\\\underline{c = \sqrt{7}}\\\\\\P = \frac{1}{2}ah = \frac{1}{2}(\sqrt{2}+\sqrt{3})\cdot2 =\underline{ \sqrt{2}+\sqrt{3}}[/tex]
Obliczamy promień okręgu opisanego na tym trójkącie:
[tex]R = \frac{abc}{4P}\\\\R = \frac{(\sqrt{2}+\sqrt{3})\cdot\sqrt{6}\cdot\sqrt{7}}{4(\sqrt{2}+\sqrt{3})}=\frac{\sqrt{42}}{4}}[/tex]
Obliczamy sinus kąta ACB:
[tex]sin\alpha = \frac{a}{2R}\\\\sin\alpha = \frac{\sqrt{2}+\sqrt{3}}{2\cdot\frac{\sqrt{42}}{4}}=\frac{2(\sqrt{2}+\sqrt{3})}{\sqrt{42}}\cdot\frac{\sqrt{42}}{\sqrt{42}} = \frac{2\sqrt{42}(\sqrt{2}+\sqrt{3})}{42} =\frac{\sqrt{42}(\sqrt{2}+\sqrt{3})}{21}\approx0,971[/tex]
