b
(b)For equilibrium of the system torques on \(M_1\) and \(M_2\) due to \(B_H\) must counter balance each other 

i.e. \({M_1} \times {B_H} = {M_2} \times {B_H}\). If \(\theta\) is the angle between \(M_1\) and \(B_H\) will be \((90 - \theta )\); 

so \({M_1}{B_H}\sin \theta = {M_2}{B_H}\sin (90 - \theta )\)

\( \Rightarrow \tan \theta = \frac{{{M_2}}}{{{M_1}}} = \frac{M}{{3M}} = \frac{1}{3} \Rightarrow \theta = {\tan ^{ - 1}}\left( {\frac{1}{3}} \right)\)