Difference between revisions of "1981 AHSME Problems/Problem 21"
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− | Problem 21 | + | ==Problem 21== |
In a triangle with sides of lengths <math>a</math>, <math>b</math>, and <math>c</math>, <math>(a+b+c)(a+b-c) = 3ab</math>. The measure of the angle opposite the side length <math>c</math> is | In a triangle with sides of lengths <math>a</math>, <math>b</math>, and <math>c</math>, <math>(a+b+c)(a+b-c) = 3ab</math>. The measure of the angle opposite the side length <math>c</math> is | ||
<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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+ | 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>. |
Revision as of 15:44, 6 March 2020
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 .