A block of mass $1 \mathrm{kg}$ is at rest relative to a smooth wedge moving leftwards with constant acceleration $a=5{ \mathrm{ms}}^{-2}$. Let  $N$ be the normal reaction between the block and the wedge. Then, $g= 10m{s}^{-2}.$

Given block of mass $1 \mathrm{kg}$ at rest on the smooth inclined plane
Wedge has an acceleration of $5 \mathrm{m}/{\mathrm{s}}^2$ so a pseudo-force of $5 \mathrm{N}$ will act on block
And its direction will be Negative to the direction of a
So the block has these three forces acting on it
$\mathrm{mg}=10 \mathrm{N}$ (downward), Normal (at an angle \theta from vertical) and $5 \mathrm{N}$ downward
The normal force will be the resultant of $mg$ and pseudo-force
$=\sqrt{{10}^2+5^2}$
$=5\sqrt{5}$
We have $\tan \theta =\mathrm{ma}/\mathrm{mg}$
$=5 / 10 =1 / 2$