# Difference between revisions of "1981 AHSME Problems/Problem 21"

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− | First notice that exchanging <math>a</math> for <math>b</math> in the original equation must also work. Therefore, <math>a=b</math>. Replacing <math>b</math> for <math>a</math> and expanding/simplifying in the original equation yields <math>4a^2-c^2=3a^2</math>, or <math>a^2=c^2</math>. Since <math>a</math> and <math>c</math> are positive, <math>a=c</math>. Therefore, we have an equilateral triangle and the angle opposite <math>c</math> is just <math>\fbox | + | First notice that exchanging <math>a</math> for <math>b</math> in the original equation must also work. Therefore, <math>a=b</math>. Replacing <math>b</math> for <math>a</math> and expanding/simplifying in the original equation yields <math>4a^2-c^2=3a^2</math>, or <math>a^2=c^2</math>. Since <math>a</math> and <math>c</math> are positive, <math>a=c</math>. Therefore, we have an equilateral triangle and the angle opposite <math>c</math> is just <math>\textbf\fbox{(D)}\ 60^\circ\qquad</math>. |

## Revision as of 15:41, 6 March 2020

## Problem 21

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

## Solution

First notice that exchanging for in the original equation must also work. Therefore, . 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 $\textbf\fbox{(D)}\ 60^\circ\qquad$ (Error compiling LaTeX. ! Argument of \fbox has an extra }.).