Синусова и косинусова теорема

[Гост]

Given a triangle ABC, the shortest side and the circumradius are in the ratio 6:5. The other two sides are 20 cm and 21 cm. Find the shortest side of the triangle.

[math10.com]

Lets [tex]BC=a=20 ; AC=b=21 ; AB=c ; \Right \frac{c}{R}=\frac{6}{5}[/tex]

[tex]c<a<b \Right \gamma <\alpha <\beta \Right \gamma <90^\circ[/tex]

From Sin Th [tex]\Right \frac{c}{sin \gamma}=2R \Right sin \gamma =\frac{c}{2R}=\frac{3}{5}[/tex]

[tex]cos\gamma =\sqrt{1-sin^2\gamma}=\sqrt{1-\frac{9}{25}}=\frac{4}{5}[/tex]

From Cos Th [tex]\Right c^2=a^2+b^2-2.a.b.cos \gamma=20^2+21^2-2.20.21.\frac{4}{5}=400+441-672=169 \Right c=13[/tex]