# Difference between revisions of "2011 AMC 10B Problems/Problem 7"

## Problem

The sum of two angles of a triangle is $\frac{6}{5}$ of a right angle, and one of these two angles is $30^{\circ}$ larger than the other. What is the degree measure of the largest angle in the triangle?

$\textbf{(A)}\ 69 \qquad\textbf{(B)}\ 72 \qquad\textbf{(C)}\ 90 \qquad\textbf{(D)}\ 102 \qquad\textbf{(E)}\ 108$

## Solution

The sum of two angles in a triangle is $\frac{6}{5}$ of a right angle $\longrightarrow \frac{6}{5} \times 90 = 108$

If $x$ is the measure of the first angle, then the measure of the second angle is $x+30$. $$x + x + 30 = 108 \longrightarrow 2x = 78 \longrightarrow x = 39$$

Now we know the measure of two angles are $39^{\circ}$ and $69^{\circ}$. By the Triangle Sum Theorem, the sum of all angles in a triangle is $180^{\circ},$ so the final angle is $72^{\circ}$. Therefore, the largest angle in the triangle is $\boxed{\mathrm{(B) \ } 72}$