Đáp án:
$\begin{array}{l}
a)\widehat A = {30^0}\\
\Rightarrow \widehat B = {90^0} - \widehat A = {60^0}\\
+ )tan\widehat A = \dfrac{a}{b}\\
\Rightarrow \tan {30^0} = \dfrac{a}{{10}}\\
\Rightarrow a = 10.\dfrac{1}{{\sqrt 3 }} = \dfrac{{10\sqrt 3 }}{3}\left( {cm} \right)\\
\Rightarrow {c^2} = {a^2} + {b^2} = \dfrac{{400}}{3}\\
\Rightarrow c = \dfrac{{20\sqrt 3 }}{3}\left( {cm} \right)\\
Vay\,a = \dfrac{{10\sqrt 3 }}{3};c = \dfrac{{20\sqrt 3 }}{3};\widehat B = {60^0}\\
b)c = 20;\widehat B = {35^0}\\
\Rightarrow \widehat A = {90^0} - {35^0} = {55^0}\\
+ \sin \widehat A = \dfrac{a}{c}\\
\Rightarrow \sin {55^0} = \dfrac{a}{{20}}\\
\Rightarrow a = 16,38\\
\Rightarrow b = \sqrt {{c^2} - {a^2}} = 11,47\\
c)a = 21;b = 18\\
\Rightarrow c = \sqrt {{a^2} + {b^2}} = 3\sqrt {85} \\
+ )\sin \widehat A = \dfrac{a}{c} = \dfrac{{21}}{{3\sqrt {85} }} = \dfrac{7}{{\sqrt {85} }}\\
\Rightarrow \widehat A = {50^0}\\
\Rightarrow \widehat B = {40^0}\\
d)a = 82;\widehat A = {42^0}\\
\Rightarrow \widehat B = {48^0}\\
\sin \widehat A = \dfrac{a}{c}\\
\Rightarrow \sin {42^0} = \dfrac{{82}}{c}\\
\Rightarrow c = 122,54\\
\Rightarrow b = \sqrt {{c^2} - {a^2}} = 91
\end{array}$
