Cho tam giác ABC có $\angle A = {60^0},\,\,AB = 4,\,\,AC = 6.$  Cạnh BC  bằng:

$R = \dfrac{a}{{\sin A}}$

$R = \dfrac{{\sqrt 2 }}{2}b$

$R = \dfrac{{\sqrt 2 }}{2}c$

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

$S = \dfrac{1}{2}ca$

$S = \dfrac{{ - \sqrt 2 }}{4}ac$

 $S = \dfrac{{\sqrt 2 }}{4}bc$

$S = \dfrac{{\sqrt 2 }}{4}ca$

${a^2} = {b^2} + {c^2} + \sqrt 2 ab.$

$\dfrac{b}{{\sin A}} = \dfrac{a}{{\sin B}}$

$\sin B = \dfrac{{ - \sqrt 2 }}{2}$

${b^2} = {c^2} + {a^2} - 2ca\cos {135^o}.$

$\sin A = \sin \,(B + C)$

$\cos A = \cos \,(B + C)$

$\;\cos A > 0$

$\sin A\,\, \le 0$

$S = \dfrac{{abc}}{{4r}}$

$r = \dfrac{{2S}}{{a + b + c}}$

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

$S = r\,(a + b + c)$

13,114

12,104

11,154

10,151