2008 AMC 10B Problem 14

Find where $A$ is first, then turn it.

Finding $A$.

1. The right angle is at $B = (5, 0)$, so $AB$ goes straight up from $B$. That means $A = (5, h)$, where $h$ is the length $AB$.
2. Triangle $OBA$ has angles $30^\circ$ at $O$, $90^\circ$ at $B$ and $60^\circ$ at $A$. In a $30$-$60$-$90$ triangle, the side across from $60^\circ$ is $\sqrt3$ times the side across from $30^\circ$.
3. Across from the $60^\circ$ angle (at $A$) is $OB = 5$. Across from the $30^\circ$ angle (at $O$) is $AB = h$. So
$$5 = \sqrt3\, h, \qquad h = \frac{5}{\sqrt3} = \frac{5\sqrt3}{3}.$$
4. So $A = \left(5, \tfrac{5}{3}\sqrt3\right)$.

Turning $A$ by $90^\circ$ counterclockwise.

1. Picture a simple point first: $(1, 0)$ turned a quarter turn counterclockwise lands at $(0, 1)$, and $(0, 1)$ lands at $(-1, 0)$. In general, $(x, y)$ goes to $(-y, x)$: swap the coordinates, then make the new first one negative.
2. For $A$: $\left(5, \tfrac{5}{3}\sqrt3\right)$ goes to $\left(-\tfrac{5}{3}\sqrt3, 5\right)$.
3. Check: $A$ was in the upper right, and a quarter turn counterclockwise should move it to the upper left. The new point has a negative first coordinate and a positive second one, as expected.

The answer is $\boxed{\textbf{(B)}\ \left(-\frac{5}{3}\sqrt{3}, 5\right)}$.

Break the problem into "where is the point?" and "where does the turn send it?". The $30$-$60$-$90$ side ratios place $A$ without any trigonometry. The quarter-turn rule $(x, y) \to (-y, x)$ is worth remembering, and a quick check of which quadrant the image lands in catches sign mistakes.