Nếu tam giác $ABC$  có \({a^2} < {b^2} + {c^2}\) thì

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

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

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

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

\(a = 2R\cos A\)          

\(a = 2R\sin A\)

\(a = 2R\tan A\)

\(a = R\sin A\)

\(a\sin B = b\sin A\)

\(a\sin A = b\sin B\)

\(a\cos B = b\cos A\)

\(a\cos A = b\cos B\)

\(bc = 2R.{h_a}\)

\(ac = R.{h_b}\)

\({a^2} = R.{h_a}\)

\(ab = 4R.{h_c}\)

\({h_a} = b\sin C\)

\({h_a} = c\sin B\)      

\({h_b} = b\sin B\)     

\(c{h_c} = ab\sin C\)

$10$

\(\dfrac{{5\sqrt 6 }}{2}\)                  

\(5\sqrt 3 \)      

\(5\sqrt 2 \)

\(2\sqrt {129} \)          

\(14\)

\(98\)

\(2\sqrt {69} \)