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

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We will try to solve for a possible value of the variables. First notice that exchanging <math>a</math> for <math>b</math> in the original equation must also work. Therefore, <math>a=b</math> works. 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{(D)}\ 60^\circ\qquad</math>. | We will try to solve for a possible value of the variables. First notice that exchanging <math>a</math> for <math>b</math> in the original equation must also work. Therefore, <math>a=b</math> works. 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{(D)}\ 60^\circ\qquad</math>. | ||

+ | |||

+ | ==Solution 2== | ||

+ | |||

+ | <cmath>(a+b+c)(a+b-c)=3ab</cmath> | ||

+ | <cmath>a^2+2ab+b^2-c^2=3ab</cmath> | ||

+ | <cmath>a^2+b^2-c^2=ab</cmath> | ||

+ | <cmath>c^2=a^2+b^2-ab</cmath> | ||

+ | This looks a lot like Law of Cosines, which is <math>c^2=a^2+b^2-2ab\cosc</math> | ||

+ | <cmath>c^2=a^2+b^2-ab=a^2+b^2-2ab\cosc</cmath> | ||

+ | <cmath>ab=2ab\cosc</cmath> | ||

+ | <cmath>\frac{1}{2}=\cosc</cmath> | ||

+ | <math>\cosc</math> is <math>\frac{1}{2}, so the angle opposite side </math>c<math> is </math>\boxed{60^\circ}$ | ||

+ | |||

+ | -aopspandy |

## Revision as of 19:19, 18 June 2021

## 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