Đáp án: $\begin{array}{l}
a){S_{ABC}} = 6\sqrt 3 \\
b)R = \dfrac{{7\sqrt 3 }}{3}\\
c))A = {98^0}25';B = {60^0};C = {21^0}{75^0}
\end{array}$
Giải thích các bước giải:
$\begin{array}{l}
AB = 3;AC = 7;BC = 8\\
\Leftrightarrow c = 3;b = 7;a = 8\\
\Leftrightarrow p = \dfrac{{a + b + c}}{2} = \dfrac{{3 + 7 + 8}}{2} = 9\\
a)\\
{S_{ABC}} = \sqrt {p.\left( {p - a} \right).\left( {p - b} \right).\left( {p - c} \right)} \\
= \sqrt {9.\left( {9 - 8} \right).\left( {9 - 7} \right).\left( {9 - 3} \right)} \\
= \sqrt {9.2.6} \\
= 6\sqrt 3 \\
b)\\
Do:{S_{ABC}} = \dfrac{{a.b.c}}{{4R}}\\
\Leftrightarrow R = \dfrac{{abc}}{{4{S_{ABC}}}} = \dfrac{{3.7.8}}{{4.6\sqrt 3 }} = \dfrac{{7\sqrt 3 }}{3}\\
Vay\,R = \dfrac{{7\sqrt 3 }}{3}\\
c)\\
\cos A = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}} = \dfrac{{{7^2} + {3^2} - {8^2}}}{{2.7.3}} = \dfrac{{ - 1}}{7}\\
\Leftrightarrow A = {98^0}25'\\
\cos B = \dfrac{{{a^2} + {c^2} - {b^2}}}{{2ac}} = \dfrac{{{8^2} + {3^2} - {7^2}}}{{2.8.3}} = \dfrac{1}{2}\\
\Leftrightarrow B = {60^0}\\
\Leftrightarrow C = {180^0} - A - B = {21^0}{75^0}\\
Vay\,A = {98^0}25';B = {60^0};C = {21^0}{75^0}
\end{array}$
