Đáp án:
$\begin{array}{l}
a)Theo\,Pytago:\\
{b^2} + {c^2} = {a^2}\\
\Rightarrow {c^2} = {18^2} - {8^2} = 260\\
\Rightarrow c = 2\sqrt {65} \\
\Rightarrow \tan \widehat C = \dfrac{c}{b} = \dfrac{{2\sqrt {65} }}{8} = \dfrac{{\sqrt {65} }}{4}\\
\Rightarrow \widehat C = \arctan \dfrac{{\sqrt {65} }}{4}\\
\Rightarrow \widehat C = {90^0} - \arctan \dfrac{{\sqrt {65} }}{4}\\
b)b = 20;\widehat C = {38^0}\\
\Rightarrow \widehat B = {52^0}\\
\Rightarrow \tan \widehat C = \dfrac{c}{b}\\
\Rightarrow c = \tan {38^0}.20 = 15,6\\
\Rightarrow a = \sqrt {{b^2} + {c^2}} = \sqrt {643,36} = 25,36\\
c)\tan \widehat B = \dfrac{3}{4};c = 4\\
\Rightarrow \widehat B = {36^0}\\
\Rightarrow \widehat C = {54^0}\\
\Rightarrow \dfrac{3}{4} = \dfrac{b}{c}\\
\Rightarrow b = 3\\
\Rightarrow a = \sqrt {{b^2} + {c^2}} = \sqrt {{3^2} + {4^2}} = 5
\end{array}$
