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

(Created page with "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 sid...")
 
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Problem 21
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==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>

Revision as of 15:33, 6 March 2020

Problem 21

In a triangle with sides of lengths $a$, $b$, and $c$, $(a+b+c)(a+b-c) = 3ab$. The measure of the angle opposite the side length $c$ is

$\textbf{(A)}\ 15^\circ\qquad\textbf{(B)}\ 30^\circ\qquad\textbf{(C)}\ 45^\circ\qquad\textbf{(D)}\ 60^\circ\qquad\textbf{(E)}\ 150^\circ$

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