### Instruction: The point $(4 + 7 \sqrt{3}, 7 - 4 \sqrt{3})$ is rotated $60^\circ$ counter-clockwise about the origin.  Find the resulting point.### Response: I know that to rotate a point $(x, y)$ counter-clockwise by an angle $\theta$ about the origin, I can use the formula $(x \cos \theta - y \sin \theta, x \sin \theta + y \cos \theta)$.
In this case, $\theta = 60^\circ$, so I need to plug in the values of $x$ and $y$ from the given point, and the values of $\sin 60^\circ$ and $\cos 60^\circ$.
I recall that $\sin 60^\circ = \frac{\sqrt{3}}{2}$ and $\cos 60^\circ = \frac{1}{2}$.
So, the rotated point is $(4 + 7 \sqrt{3}) \frac{1}{2} - (7 - 4 \sqrt{3}) \frac{\sqrt{3}}{2}, (4 + 7 \sqrt{3}) \frac{\sqrt{3}}{2} + (7 - 4 \sqrt{3}) \frac{1}{2})$.
I need to simplify this expression by multiplying out the brackets and combining like terms.
I get $(2 + \frac{7 \sqrt{3}}{2} - \frac{7 \sqrt{3}}{2} + 6, 2 \sqrt{3} + \frac{21}{2} + \frac{7}{2} - 2 \sqrt{3})$.
I see that the terms with $\sqrt{3}$ cancel out in the first coordinate, and in the second coordinate.
I am left with $(8, 14)$.