Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
BC = a = \sqrt {{b^2} + {c^2} - 2bc.\cos A} = \sqrt {{{20}^2} + {{35}^2} - 2.20.35.\cos 60^\circ } = 5\sqrt {37} \left( {cm} \right)\\
b,\\
{S_{ABC}} = \dfrac{1}{2}bc.\sin A = \dfrac{1}{2}.20.35.\sin 60^\circ = 175\sqrt 3 \,\,\,\,\,\left( {c{m^2}} \right)\\
c,\\
\cos B = \dfrac{{{a^2} + {c^2} - {b^2}}}{{2ac}} = \dfrac{{{{\left( {5\sqrt {37} } \right)}^2} + {{35}^2} - {{20}^2}}}{{2.5\sqrt {37} .35}} > 0\\
\Rightarrow 0^\circ < \widehat B < 90^\circ \\
d,\\
AH = \dfrac{{2{S_{ABC}}}}{{BC}} = \dfrac{{2.175\sqrt 3 }}{{5\sqrt {37} }} = \dfrac{{70\sqrt {111} }}{{37}}\,\,\,\,\,\left( {cm} \right)\\
e,\\
{S_{ABC}} = \dfrac{{abc}}{{4R}} \Rightarrow R = \dfrac{{abc}}{{4{S_{ABC}}}} = \dfrac{{5\sqrt {37} .20.35}}{{4.175\sqrt 3 }} = \dfrac{{5\sqrt {111} }}{3}\,\,\,\,\left( {cm} \right)\\
{S_{ABC}} = \dfrac{{a + b + c}}{2}.r \Rightarrow r = \dfrac{{2{S_{ABC}}}}{{a + b + c}} = \dfrac{{2.175\sqrt 3 }}{{20 + 35 + 5\sqrt {37} }} = \dfrac{{350\sqrt 3 }}{{55 + 5\sqrt {37} }} = \dfrac{{70\sqrt 3 }}{{11 + \sqrt {37} }}\,\,\,\,\left( {cm} \right)
\end{array}\)
