Determine Coordinates of $30^\circ$ Counterclockwise Rotation

Determine the coordinates of $$A'$$, the image of $$A$$ after a $$30^\circ$$ counterclockwise rotation about $$P\left(3,2\right)$$

I have figured out that $$\overline{PA}=5$$ and that a $$36.87^\circ$$ counterclockwise rotation maps $$A'$$ onto $$(3,7)$$, but can't seem to find a way to progress.

• Can you rotate the point (3,4) about the origin by 30 degrees? Commented Mar 14 at 20:50

If $$\alpha$$ is the angle by which you have to rotate $$(8,2)$$ about $$P$$ in order to get $$A$$, then $$A=(3+5\cos\alpha,\;2+5\sin\alpha)$$ and $$\cos\alpha = \frac{3}{5}$$ and $$\sin\alpha=\frac{4}{5}.$$ You get $$A'$$ by simply rotating $$30^{\circ}$$ more: $$A'=(3+5\cos(\alpha+30^{\circ}),\;2+5\sin(\alpha+30^{\circ})\;)$$ Now use $$\cos(\alpha+\beta) = \cos\alpha\,\cos\beta-\sin\alpha\,\sin\beta \\ \sin(\alpha+\beta) = \sin\alpha\,\cos\beta+\cos\alpha\,\sin\beta$$ and $$\cos 30^{\circ} = \frac{\sqrt{3}}{2} \\ \sin 30^{\circ} = \frac{1}{2}$$ to get $$A'=\left( 3+5\cdot\frac{3}{5}\cdot\frac{\sqrt{3}}{2}-5\cdot\frac{4}{5}\cdot\frac{1}{2} \;,\; 2+5\cdot\frac{4}{5}\cdot\frac{\sqrt{3}}{2}+5\cdot\frac{3}{5}\cdot\frac{1}{2} \right) \\ = \left( 1+\frac{3\sqrt{3}}{2}\;,\; \frac{7}{2}+2\sqrt{3} \right)$$