Đáp án:
$\begin{array}{l}
Gia\,su:a = 1\\
\Rightarrow AB = 1;AC = \sqrt 3 \\
Do:A{C^2} = A{B^2} + B{C^2} - 2.AB.BC.\cos B\\
\Rightarrow {\left( {\sqrt 3 } \right)^2} = {1^2} + B{C^2} - 2.BC.\cos {60^0}\\
\Rightarrow B{C^2} - BC - 2 = 0\\
\Rightarrow \left( {BC - 2} \right)\left( {BC + 1} \right) = 0\\
\Rightarrow BC = 2\\
hay\,BC = 2a\\
\Rightarrow {S_{ABC}} = \dfrac{1}{2}.BA.BC.sinB\\
= \dfrac{1}{2}.a.2a.\sin {60^0}\\
= \dfrac{{\sqrt 3 {a^2}}}{2}\\
Do:{S_{ABC}} = \dfrac{{AB.AC.BC}}{{4R}} = p.r\\
\Rightarrow \left\{ \begin{array}{l}
R = \dfrac{{a.a\sqrt 3 .2a}}{{4S}} = \dfrac{{2\sqrt 3 {a^3}}}{{2\sqrt 3 {a^2}}} = a\\
r = \dfrac{{2S}}{{AB + AC + BC}} = \dfrac{{\sqrt 3 {a^2}}}{{3a + a\sqrt 3 }} = \dfrac{a}{{\sqrt 3 + 1}}
\end{array} \right.
\end{array}$
