The resultant of two vectors A and B is perpendicular to vector A and its magnitude is equal to half of the magnitude of vector B, The angle between A and B is

Here $(\mathrm{A}+\mathrm{B})\cdot \mathrm{A}=0$
$\begin{array}{l}{\mathrm{A}}^2+\mathrm{ABcos}\mathrm{\theta }=0\\ \mathrm{or} \mathrm{ABcos}\mathrm{\theta }=-{\mathrm{A}}^2....\left(1\right)\\ \mathrm{Bcos}\mathrm{\theta }=-\mathrm{A}...\left(2\right)\end{array}$
Further, $|\mathrm{A}+\mathrm{B}|=\frac{1}{2}\left|\mathrm{B}\right|$
$\begin{array}{l}\mathrm{or} (\mathrm{A}+\mathrm{B})\cdot (\mathrm{A}+\mathrm{B})={\left(\frac{1}{2}\left|\mathrm{B}\right|\right)}^2\\ {\mathrm{A}}^2+{\mathrm{B}}^2+2\mathrm{ABcos}\mathrm{\theta }=\frac{1}{4}{\mathrm{B}}^2\\ \mathrm{or} \left({\mathrm{A}}^2+\mathrm{ABcos}\mathrm{\theta }\right)+{\mathrm{B}}^2+\mathrm{ABcos}\mathrm{\theta }=\frac{1}{4}{\mathrm{B}}^2\\ 0+{\mathrm{B}}^2+\mathrm{ABcos}\mathrm{\theta }=-\frac{1}{4}{\mathrm{B}}^2\end{array}$
$\mathrm{or} \mathrm{ABcos}\mathrm{\theta }=-\frac{3}{4}{\mathrm{B}}^2\dots \left(3\right)$
From eqs. (1) and (3)
${\mathrm{A}}^2=\frac{3}{4}{\mathrm{B}}^2 \mathrm{or} \mathrm{A}=(\sqrt{3}/2)\mathrm{B}$
From eq. (2), $\mathrm{Bcos}\mathrm{\theta }=-(\sqrt{3}/2)\mathrm{B}$
or $\cos \mathrm{\theta }=-\sqrt{3}/2 \mathrm{or} \mathrm{\theta }={150}^{\circ }$