In figure, the acceleration of A is ${\overrightarrow{\mathrm{a}}}_{\mathrm{A}}=15\hat{\mathrm{i}}+15\hat{\mathrm{j}}$. Then the acceleration of B is (A remains in contact with B)

From constraint, the acceleration of both block and wedge should be same in a direction perpendicular to the inclined plane as shown in figure.

${\left({\mathrm{a}}_{\mathrm{A}}\right)}_{\perp }={\left({\mathrm{a}}_{\mathrm{B}}\right)}_{\perp }$,

${\mathrm{a}}_{\mathrm{AX}}=15,{\mathrm{a}}_{\mathrm{AY}}=15$

${\mathrm{a}}_{\mathrm{AX}}\cos {53}^{\circ }-{\mathrm{a}}_{\mathrm{AY}}\cos {37}^{\circ }={\mathrm{a}}_{\mathrm{B}}\cos {53}^{\circ }$

or ${\mathrm{a}}_{\mathrm{B}}=-5{\mathrm{ms}}^{-1}\text{ or }{\overrightarrow{\mathrm{a}}}_{\mathrm{B}}=-5\hat{\mathrm{i}}$