Difference between revisions of "1981 AHSME Problems/Problem 21"
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<math>\textbf{(A)}\ 15^\circ\qquad\textbf{(B)}\ 30^\circ\qquad\textbf{(C)}\ 45^\circ\qquad\textbf{(D)}\ 60^\circ\qquad\textbf{(E)}\ 150^\circ</math> | <math>\textbf{(A)}\ 15^\circ\qquad\textbf{(B)}\ 30^\circ\qquad\textbf{(C)}\ 45^\circ\qquad\textbf{(D)}\ 60^\circ\qquad\textbf{(E)}\ 150^\circ</math> | ||
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| + | ==Solution== | ||
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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 45^\circ\qquad\textbf{(D)}\. | ||
Revision as of 14:38, 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 45^\circ\qquad\textbf{(D)}\.
