Giải thích các bước giải:
Bài 3:
Ta có:
\(\begin{array}{l}
{S_{ABC}} = \frac{1}{2}.BA.BC.\sin B\\
\Leftrightarrow 3\sqrt 3 = \frac{1}{2}.3.4.\sin B\\
\Leftrightarrow \sin B = \frac{{\sqrt 3 }}{2}\\
\cos B = \pm \sqrt {1 - {{\sin }^2}B} = \pm \frac{1}{2}\\
TH1:\,\,\,\,\,\,cosB = \frac{1}{2}\\
\Rightarrow AC = \sqrt {A{B^2} + B{C^2} - 2.\cos B.AB.BC} = \sqrt {13} \\
TH2:\,\,\,\,\,\cos B = - \frac{1}{2}\\
\Rightarrow AC = \sqrt {A{B^2} + B{C^2} - 2.\cos B.AB.BC} = \sqrt {37}
\end{array}\)
Bài 4:
\(\begin{array}{l}
a,\\
{a^2} = {b^2} + {c^2} - 2.b.c.\cos A\\
\Leftrightarrow {a^2} = {5^2} + {8^2} - 2.5.8.\cos 120^\circ \\
\Rightarrow a = \sqrt {129} \\
\frac{a}{{\sin A}} = \frac{b}{{\sin B}} = \frac{c}{{\sin C}} = 2R\\
\Leftrightarrow \frac{{\sqrt {129} }}{{\sin 120^\circ }} = \frac{8}{{\sin B}} = \frac{5}{{\sin C}} = 2R\\
\Rightarrow \left\{ \begin{array}{l}
\widehat B = 37,59^\circ \\
\widehat C = 22,41^\circ
\end{array} \right.\\
b,\\
{S_{ABC}} = \frac{1}{2}.b.c.\sin A = \frac{1}{2}.5.8.sin120^\circ = 10\sqrt 3
\end{array}\)
Bài 5:
Tam giác ABC cân tại A nên \(\widehat B = \widehat C = 30^\circ \Rightarrow \widehat A = 180^\circ - \widehat B - \widehat C = 120^\circ \)
Ta có:
\(\begin{array}{l}
\frac{a}{{\sin A}} = \frac{b}{{\sin B}} = \frac{c}{{\sin C}} = 2R\\
\Leftrightarrow \frac{5}{{\sin 120^\circ }} = \frac{b}{{\sin 30^\circ }} = \frac{c}{{\sin 30^\circ }} = 2R\\
\Rightarrow \left\{ \begin{array}{l}
b = c = \frac{{5\sqrt 3 }}{3}\\
R = \frac{{5\sqrt 3 }}{3}
\end{array} \right.\\
{S_{ABC}} = \frac{{a.b.c}}{{4R}} = \frac{{25\sqrt 3 }}{{12}}
\end{array}\)
