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

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

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

− | <cmath>c^2=a^2+b^2-ab=a^2+b^2-2ab\ | + | <cmath>c^2=a^2+b^2-ab=a^2+b^2-2ab\cos{c}</cmath> |

− | <cmath>ab=2ab\ | + | <cmath>ab=2ab\cos{c}</cmath> |

− | <cmath>\frac{1}{2}=\ | + | <cmath>\frac{1}{2}=\cos{c}</cmath> |

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

-aopspandy | -aopspandy |

## Revision as of 19:20, 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 . is , so the angle opposite side is

-aopspandy