?

\[\]
eqalign{

& T = g{m_A} \cr

& T = 20m{\text{\ N}}\ \ \cr \ \ &\ \ \ \cr \ \ &\ {F_R}\ =\ T\ -\ P\ -\ fat\ \ \cr \ \ &\ m\ \cdot\ a\ =\ 20m\ -\ mg\sin\ \theta\ \ -\ \mu\ mg\sin\ \theta\ \ \ \cr \ \ &\ m\ \cdot\ 0\ =\ 20m\ -\ 20\sin\ \theta\ \ -\ 20m\mu\ \sin\ \theta\ \ \ \cr \ \ &\ 0\ =\ 20m\ \cr }\ \ \ Portanto,\ \ obtemos\ que:\ $\boxed{\sin\ \theta\ \ =\ \dfrac{1}{{2\mu