Tam giác $ABC$  có $BC = 10$  và \(\widehat A = {30^0}\). Khi đó, bán kính đường tròn ngoại tiếp tam giác $ABC$  là:

$\sin {743^0} = \sin {23^0}$

$\sin {743^0} =  - \sin {23^0}$         

$\sin {743^0} = \cos {\rm{2}}{{\rm{3}}^0}$         

$\sin {743^0} =  - \cos {\rm{2}}{{\rm{3}}^0}$

\({a^2} = {b^2} + {c^2} - bc\cos A\)           

\({a^2} = {b^2} + {c^2} + bc\cos A\)

${a^2} = {b^2} + {c^2} - 2bc\cos A$          

\({a^2} = {b^2} + {c^2} + 2bc\cos A\)

\(a = 2R\cos A\)          

\(a = 2R\sin A\)

\(a = 2R\tan A\)

\(a = R\sin A\)

$\sin x > 0$    

$\cos x < 0$

$\tan x > 0$

$\cot x > 0$

\(\cot \alpha \tan \alpha  = 1,\quad \alpha  \ne \dfrac{{k\pi }}{2},k \in Z\)   

$1 + {\tan ^2}\alpha  = \dfrac{1}{{{{\cos }^2}\alpha }},{\rm{ }}\alpha  \ne {\rm{k}}\pi {\rm{, k}} \in Z$

${\sin ^2}\alpha  + c{\rm{o}}{{\rm{s}}^2}\beta  = 1$

$1 + {\cot ^2}\alpha  = \dfrac{1}{{{{\sin }^2}\alpha }},{\rm{ }}\alpha  \ne \dfrac{\pi }{2}{\rm{ + k}}\pi {\rm{, k}} \in Z$

$n-p$

$m + p$

$m-p$

$n + p$

$0$     

$1$     

$2$                 

$4$