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> | ||
− | ==Solution== | + | ==Solution 1== |
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>. | 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>. | ||
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-aopspandy | -aopspandy | ||
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+ | ==See also== | ||
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+ | {{AHSME box|year=1981|num-b=20|num-a=22}} | ||
+ | {{MAA Notice}} |
Latest revision as of 14:13, 28 June 2025
Contents
Problem 21
In a triangle with sides of lengths ,
, and
,
. The measure of the angle opposite the side length
is
Solution 1
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
See also
1981 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.