Let $\overrightarrow{A}=\hat{i}A\cos \theta +\hat{j}Asin\theta$  be any vector. Another vector $\overrightarrow{B}$  which is perpendicular to $\overrightarrow{A}$  can be expressed as : 

$\overrightarrow{A}=\hat{i}A\cos \theta +\hat{j}A\sin \theta$ 

For $\overrightarrow{B}tobe\perp to\overrightarrow{A}$

      $\overrightarrow{B}.\overrightarrow{A}=BA\cos {90}^0=0$

       $\overrightarrow{B}.\overrightarrow{A}=\left(\hat{i}B\sin \theta -\hat{j}B\cos \theta \right)$ .$\left(\hat{i}A\cos \theta -\hat{j}Asin\theta \right)$

              $=AB\sin \theta \cos \theta -BA\sin \theta \cos \theta =0$

Hence, $\overrightarrow{B}=\hat{i}B\sin \theta -\hat{j}B\cos \theta$