Đáp án:
$\begin{array}{l}
a)\sin \beta = \dfrac{1}{4}\\
Do:0 < \beta < {90^0}\\
\Rightarrow \cos \beta > 0\\
\Rightarrow \cos \beta = \sqrt {1 - {{\sin }^2}\beta } \\
= \sqrt {1 - \dfrac{1}{{16}}} = \dfrac{{\sqrt {15} }}{4}\\
\Rightarrow \left\{ \begin{array}{l}
\tan \beta = \dfrac{{\sin \beta }}{{\cos \beta }} = \dfrac{{\dfrac{1}{4}}}{{\dfrac{{\sqrt {15} }}{4}}} = \dfrac{{\sqrt {15} }}{{15}}\\
\cot \beta = \sqrt {15}
\end{array} \right.\\
b){90^0} < a < {180^0}\\
\Rightarrow \sin a < 0\\
\Rightarrow \sin a = - \sqrt {1 - {{\cos }^2}a} \\
= - \sqrt {1 - {{\left( {\dfrac{{ - 5}}{{13}}} \right)}^2}} \\
= - \dfrac{{12}}{{13}}\\
\Rightarrow \left\{ \begin{array}{l}
{\mathop{\rm tana}\nolimits} = \dfrac{{\sin a}}{{\cos a}} = \dfrac{5}{{12}}\\
\cot a = \dfrac{{12}}{5}
\end{array} \right.\\
c)0 < a < {180^0}\\
\Rightarrow \cos a = \pm \sqrt {1 - {{\sin }^2}a} \\
= \pm \sqrt {1 - \dfrac{{16}}{{25}}} = \pm \dfrac{3}{5}\\
\Rightarrow \left\{ \begin{array}{l}
\tan a = \pm \dfrac{{\sin a}}{{\cos a}} = \pm \dfrac{4}{3}\\
\cot a = \pm \dfrac{3}{4}
\end{array} \right.\\
d)0 < a < {90^0}\\
\Rightarrow \sin a > 0;\cos > 0\\
\Rightarrow \cot a > 0
\end{array}$
=> ko có góc a thỏa mãn đề bài.
