Cho \(\cos \alpha  =  - \frac{2}{3}\).

a) \(90^\circ  < \alpha  < 180^\circ \).

b) \(\cot \alpha \, < \,0\).                          

c) \(\sin \alpha  =  - \frac{{\sqrt 5 }}{3}\).                               

d) \(\tan \alpha  = \frac{{\sqrt 5 }}{2}\). 

a) Sai. Ta có \(90^\circ  < \alpha  < 180^\circ \) nên \(\cos \alpha  < 0\).

b) Đúng. Vì \({\sin ^2}\alpha  + {\cos ^2}\alpha  = 1 \Rightarrow {\cos ^2}\alpha  = 1 - {\sin ^2}\alpha  = 1 - {\left( {\frac{3}{5}} \right)^2} = \frac{{16}}{{25}}\).

Do đó \[\cos \alpha  =  - \sqrt {\frac{{16}}{{25}}}  =  - \frac{4}{5}\].

c) Sai. Ta có \[\tan \alpha  = \frac{{\sin \alpha }}{{\cos \alpha }} =  - \frac{3}{4} \Rightarrow \,\tan \left( {180^\circ  - \alpha } \right) =  - \tan \alpha  = \frac{3}{4}\].

d) Đúng. \[A = \frac{{\tan \alpha  - \cot \left( {180^\circ  - \alpha } \right)}}{{\sin \left( {90^\circ  - \alpha } \right)}} = \frac{{\tan \alpha  - \frac{1}{{\tan \left( {180^\circ  - \alpha } \right)}}}}{{\cos \alpha }} = \frac{{\frac{{ - 3}}{4} - \frac{4}{3}}}{{\frac{{ - 4}}{5}}} = \frac{{125}}{{48}}\].

Đáp án đúng là: B

Ta có \(\widehat C = 180^\circ  - \widehat A - \widehat B = 180^\circ  - 40^\circ  - 60^\circ  = 80^\circ \).

Áp dụng định lý sin: \[\frac{{BC}}{{\sin A}} = \frac{{AB}}{{\sin C}} \Rightarrow BC = \frac{{AB}}{{\sin C}} \cdot \sin A = \frac{5}{{\sin 80^\circ }} \cdot \sin 40^\circ  \approx 3,3\].