Given: $\overrightarrow{\mathrm{A}}=\mathrm{Acos}\mathrm{\theta }\hat{\mathrm{i}}+\mathrm{Asin}\mathrm{\theta }\hat{\mathrm{j}}$. A vector $\overrightarrow{B},$ which is perpendicular to $\overrightarrow{A}$, is given by

Clearly, $\overrightarrow{\mathrm{B}}$ should be either in the second quadrant or the fourth quadrant. In none of the given options, we have$-\hat{i}$ term. So the second quadrant is ruled out. Also $\overrightarrow{\mathrm{B}}$should make an angle of 90^o - $\mathrm{\theta }$ with the x-axis (figure). So, B should be $\overrightarrow{\mathrm{B}}$ cos (90^o - $\mathrm{\theta }$) $\hat{\mathrm{i}}-\mathrm{Bsin}\left({90}^{\circ }-\mathrm{\theta }\right)\hat{\mathrm{j}}=\mathrm{Bsin}\mathrm{\theta }\hat{\mathrm{i}}-\mathrm{Bcos}\mathrm{\theta }\hat{\mathrm{j}}$