# 1959 AHSME Problems/Problem 36

## Problem

The base of a triangle is $80$, and one side of the base angle is $60^\circ$. The sum of the lengths of the other two sides is $90$. The shortest side is: $\textbf{(A)}\ 45 \qquad\textbf{(B)}\ 40\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 17\qquad\textbf{(E)}\ 12$

## Solution

$[asy] import geometry; size(10cm); point A = (0,0); point B = (17/20,17*sqrt(3)/20); point C = (8,0); triangle ABC = triangle(A,B,C); // Triangle ABC draw(ABC); dot(A); label("A",A,SW); dot(B); label("B",B,NW); dot(C); label("C",C,SE); // Labels markscalefactor = 0.1; draw(anglemark(C,A,B)); label("60^{\circ}", A, (1.5,1.5)); label("80", midpoint(A--C), S); label("x", midpoint(A--B), NW); label("90-x", midpoint(B--C), NE); [/asy]$

Let the triangle be $\triangle ABC$ with $\measuredangle BAC=60^{\circ}$ and $AC=80$, as in the diagram. Let $AB=x$. from the problem, we know that $BC=90-x$. Now, we can apply the Law of Cosines on $\triangle ABC$ to solve for $x$: \begin{align*} (90-x)^2 &= 6400+x^2-160\cos(60^{\circ}) \\ x^2-180x+8100 &= 6400+x^2-80x \\ -100x &= -1700 \\ x &= 17 \end{align*} Because the problem asks for the shortest side of the triangle and $90-x=73>17$, our answer is $\boxed{\textbf{(D) }17}$.