# 2017 AMC 8 Problems/Problem 6

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

## Problem

If the degree measures of the angles of a triangle are in the ratio $3:3:4$, what is the degree measure of the largest angle of the triangle? $\textbf{(A) }18\qquad\textbf{(B) }36\qquad\textbf{(C) }60\qquad\textbf{(D) }72\qquad\textbf{(E) }90$

## Solution 1

The sum of the ratios is $10$. Since the sum of the angles of a triangle is $180^{\circ}$, the ratio can be scaled up to $54:54:72$ $(3\cdot 18:3\cdot 18:4\cdot 18).$ The numbers in the ratio $54:54:72$ represent the angles of the triangle. The question asks for the largest, so the answer is $\boxed{\textbf{(D) }72}$.

## Solution 2

We can denote the angles of the triangle as $3x$, $3x$, $4x$. Due to the sum of the angles in a triangle, $3x+3x+4x=180^{\circ}\implies x=18^{\circ}$. The greatest angle is $4x$ and after substitution we get $\boxed{\textbf{(D) }72}$.

~MathFun1000

## Solution 3

We know the longest side must be denoted by the 4 in the ratio. Since the ratio is 3:3:4, we know that the longest side must be $\frac{4}{3+3+4}$ of the degree total (which for all triangles is 180). Thus, $$\frac{4}{3+3+4} \cdot 180 = \frac{4}{10} \cdot 180 = \boxed{\textbf{(D) }72}$$

~Ligonmathkid2

## Solution 4 (Brute Force) NOT RECOMMENDED

Since we see the ratio is $3:3:4$, we can rule out the answer of ${\textbf{(E) }90}$ because the numbers in the ratio are too big to have $90^\circ$. Also, we are trying to find the largest angle and all the other angles except for 72 are too small to be the largest angle. Using all this, our answer is $\boxed{\textbf{(D) }72}$.

~jason.ca

## Video Solution (CREATIVE THINKING!!!)

~Education, the Study of Everything

## Video Solution

~savannahsolver

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 