2011 AMC 10B Problem 7

1. A right angle is $90^\circ$. So the two angles add up to $\frac{6}{5} \times 90 = 108$ degrees.
2. One of them is $30^\circ$ more than the other. Take away the extra $30$: $108 - 30 = 78$, which splits evenly as $39 + 39$. So the angles are $39^\circ$ and $39 + 30 = 69^\circ$.
3. The question asks about the whole triangle, so we also need the third angle. The angles of a triangle add to $180^\circ$, so the third one is $180 - 108 = 72$ degrees.
4. The three angles are $39^\circ$, $69^\circ$ and $72^\circ$. The largest is $72^\circ$.

The answer is $\boxed{\textbf{(B)}\ 72}$.

The problem tells you about only two of the angles, but asks about the largest angle of the whole triangle. Always find all three angles (using the $180^\circ$ total) before choosing the largest or smallest.

Two of the angles sum to $\frac{6}{5} \cdot 90^\circ = 108^\circ$. If they are $x$ and $x + 30$, then $2x + 30 = 108$, so $x = 39$ and the two angles are $39^\circ$ and $69^\circ$.

The third angle is $180^\circ - 108^\circ = 72^\circ$.

Comparing $39, 69, 72$, the largest angle is $\boxed{\textbf{(B)}\ 72}$.