2011 AMC 10B Problem 18

Draw rectangle $ABCD$ with $AB = 6$ along the bottom, $BC = 3$, and $M$ on $AB$. Call the angle we want $\theta$, so $\angle AMD = \angle CMD = \theta$.

1. Find a second angle equal to $\theta$. $AB$ and $DC$ are parallel, and $MD$ crosses both. So $\angle MDC = \angle AMD = \theta$ (alternate angles).
2. Triangle $MDC$ has two equal angles. Its angles at $M$ and at $D$ are both $\theta$. A triangle with two equal angles has two equal sides: the sides opposite those angles. So $CM = CD = 6$.
3. Look at triangle $MBC$. It has a right angle at $B$, with $CB = 3$ and long side $CM = 6$. The side across from $\angle CMB$ is half the long side, which is exactly what happens in a $30$-$60$-$90$ triangle. So $\angle CMB = 30^\circ$.
4. Angles along a straight line. At point $M$, the three angles $\angle AMD$, $\angle DMC$ and $\angle CMB$ fill the straight line $AB$, so they add to $180^\circ$:
$$\theta + \theta + 30 = 180, \quad\text{so}\quad 2\theta = 150 \quad\text{and}\quad \theta = 75.$$

The answer is $\boxed{\textbf{(E)}\ 75}$.

When a figure with parallel lines also says two angles are equal, look for a triangle with two equal angles; it gives you two equal sides. Here that turned an angle fact into the length $CM = 6$. Then "short side is half the long side" in a right triangle means a $30$-$60$-$90$ triangle.

Put the rectangle on a grid: $A = (0, 0)$, $B = (6, 0)$, $C = (6, 3)$, $D = (0, 3)$, $M = (m, 0)$. From step 2, $CM = 6$, so $(6 - m)^2 + 3^2 = 36$, which gives $6 - m = \sqrt{27} = 3\sqrt{3}$ and $m \approx 0.80$. In right triangle $AMD$, $\tan \theta = \frac{3}{m} \approx 3.73$, which is $\tan 75^\circ$.