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)}\.