Đáp án:
$\begin{array}{l}
a)B{C^2} = A{B^2} + A{C^2} - 2.AB.AC.\cos \widehat A\\
= {8^2} + {5^2} - 2.8.5.\cos {60^0}\\
= 49\\
\Rightarrow BC = 7\left( {cm} \right)\\
b){S_{ABC}} = \dfrac{1}{2}.AB.AC.\sin \widehat A\\
= \dfrac{1}{2}.8.5.\sin {60^0}\\
= 10\sqrt 3 \left( {c{m^2}} \right)\\
c)Do:\dfrac{{BC}}{{\sin \widehat A}} = \dfrac{{AC}}{{\sin \widehat B}}\\
\Rightarrow \sin \widehat B = 8:\dfrac{7}{{\sin {{60}^0}}} = \dfrac{{4\sqrt 3 }}{7}\\
\Rightarrow \widehat B = {82^0} < {90^0}
\end{array}$
Vậy góc B là góc nhọn
$\begin{array}{l}
d)B{M^2} = \dfrac{{A{B^2} + B{C^2}}}{2} - \dfrac{{A{C^2}}}{4}\\
= \dfrac{{{5^2} + {7^2}}}{2} - \dfrac{{{8^2}}}{4} = 21\\
\Rightarrow BM = \sqrt {21} \left( {cm} \right)\\
{S_{ABC}} = \dfrac{1}{2}.AH.BC = 10\sqrt 3 \\
\Rightarrow AH = \dfrac{{2.10\sqrt 3 }}{7} = \dfrac{{20\sqrt 3 }}{7}\left( {cm} \right)\\
e)\\
+ )S = \dfrac{{AB.AC.BC}}{{4R}}\\
\Rightarrow R = \dfrac{{AB.AC.BC}}{{4S}} = \dfrac{{5.7.8}}{{4.10\sqrt 3 }} = \dfrac{{7\sqrt 3 }}{3}\left( {cm} \right)\\
+ )p = \dfrac{{AB + AC + BC}}{2} = \dfrac{{5 + 7 + 8}}{2} = 10\\
S = p.r\\
\Rightarrow r = \dfrac{S}{p} = \dfrac{{10\sqrt 3 }}{{10}} = \sqrt 3 \left( {cm} \right)\\
Vậy\,R = \dfrac{{7\sqrt 3 }}{3}\left( {cm} \right);r = \sqrt 3 cm
\end{array}$
