# 1981 AHSME Problems/Problem 21

## Problem 21

In a triangle with sides of lengths , , and , . The measure of the angle opposite the side length is

## Solution

We will try to solve for a possible value of the variables. First notice that exchanging for in the original equation must also work. Therefore, works. Replacing for and expanding/simplifying in the original equation yields , or . Since and are positive, . Therefore, we have an equilateral triangle and the angle opposite is just .

## Solution 2

This looks a lot like Law of Cosines, which is $c^2=a^2+b^2-2ab\cosc$ (Error compiling LaTeX. ! Undefined control sequence.)

\[c^2=a^2+b^2-ab=a^2+b^2-2ab\cosc\] (Error compiling LaTeX. ! Undefined control sequence.)

\[ab=2ab\cosc\] (Error compiling LaTeX. ! Undefined control sequence.)

\[\frac{1}{2}=\cosc\] (Error compiling LaTeX. ! Undefined control sequence.)

$\cosc$ (Error compiling LaTeX. ! Undefined control sequence.) is c\boxed{60^\circ}$

-aopspandy