Cho tam giác ABCcó ˆB=60∘,ˆC=45∘, a=6. Tính b, c, R, S.

+ Tính \(\widehat A\).

Có \(\widehat A + \widehat B + \widehat C = 180^\circ \Rightarrow \widehat A = 180^\circ - \widehat B - \widehat C = 180^\circ - 60^\circ - 45^\circ = 75^\circ \)

+ Tính \(b\), \(c\), \(R\)

Áp dụng định lý sin trong \(\Delta ABC\), ta có: \(\frac{{\rm{a}}}{{\sin {\rm{A}}}} = \frac{{\rm{b}}}{{\sin {\rm{B}}}} = \frac{{\rm{c}}}{{\sin {\rm{C}}}} = 2{\rm{R}}\)

Thay số: \(\frac{6}{{\sin 75^\circ }} = \frac{{\rm{b}}}{{\sin 60^\circ }} = \frac{{\rm{c}}}{{\sin 45^\circ }} = 2{\rm{R}}\)\( \Rightarrow \left\{ {\begin{array}{*{20}{c}}{b = \frac{{6.\sin 60^\circ }}{{\sin 75^\circ }} = - 3\sqrt 6 + 9\sqrt 2 }\\{c = \frac{{6.\sin 45^\circ }}{{\sin 75^\circ }} = - 6 + 6\sqrt 3 }\\{R = \frac{6}{{2.\sin 75^\circ }} = 3\sqrt 6 - 3\sqrt 2 }\end{array}} \right.\).

+ Tính \(S\).

Có: \(S = \frac{1}{2}.{\rm{b}}.{\rm{c}}.\sin {\rm{A}} = \frac{1}{2}.\left( {{\rm{\;}} - 3\sqrt 6 + 9\sqrt 2 } \right).\left( {{\rm{\;}} - 6 + 6\sqrt 3 } \right).{\rm{\;}}\sin 75^\circ = 27 - 9\sqrt 3 \).