The resultant of two vectors $$\vec{A}$$ and $$\vec{B}$$ is perpendicular to $$\vec{A}$$ and its magnitude is half that of $$\vec{B}$$. The angle between vectors $$\vec{A}$$ and $$\vec{B}$$ is ______ °.

Let angle between $$\vec{A}$$ and $$\vec{B}$$ be $$\theta$$. Resultant $$\vec{R} = \vec{A} + \vec{B}$$ is perpendicular to $$\vec{A}$$.

$$\vec{R} \cdot \vec{A} = 0$$: $$A^2 + AB\cos\theta = 0 \implies \cos\theta = -A/B$$.

$$|\vec{R}| = B/2$$: $$A^2 + B^2 + 2AB\cos\theta = B^2/4$$.

$$A^2 + B^2 - 2A^2 = B^2/4 \implies B^2 - A^2 = B^2/4 \implies A^2 = 3B^2/4$$.

$$\cos\theta = -\frac{A}{B} = -\frac{\sqrt{3}}{2} \implies \theta = 150°$$.

The answer is $$\boxed{150}$$°.