A block of mass m is placed on a smooth inclined wedge ABC of inclination $\theta$ as shown in the figure.  The wedge is given an acceleration a towards the right.  The relation between a and $\theta$ for the block to remain at rest with respect to the wedge is 
 

According to the question, the FBD of the given condition will be
Since, the wedge is accelerating towards right with a, thus a pseudo force acts in the left direction in order to keep the block stationary. As, the system is in equilibrium,
$\therefore \sum {\text{\hspace{0.17em}F}}_x=0\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}or\hspace{0.17em}\hspace{0.17em}}\sum {\text{F}}_y=0$
$\Rightarrow \text{\hspace{0.17em}\hspace{0.17em}R}\sin \theta =\text{ma}$
or $\text{mg\hspace{0.17em}}\sin \theta =\text{ma…. (i)}$
Similarly, $\text{R}\cos \theta =\text{mg}$
$\text{or}\mathrm{mgcos}\theta =\mathrm{mg}\dots .\dots . \left(ii\right)$
Dividing Equation (i) by Equation (ii), we get
$\frac{\text{mg}\sin \theta }{\text{mg}\cos \theta }=\frac{\text{ma}}{\text{mg}}$
$\Rightarrow \tan \theta =\frac{\text{a}}{\text{g}}$
or  $\text{a}=\text{g}\tan \theta$