Difference between revisions of "1981 AHSME Problems/Problem 21"
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<cmath>a^2+b^2-c^2=ab</cmath> | <cmath>a^2+b^2-c^2=ab</cmath> | ||
<cmath>c^2=a^2+b^2-ab</cmath> | <cmath>c^2=a^2+b^2-ab</cmath> | ||
− | This looks a lot like Law of Cosines, which is <math>c^2=a^2+b^2-2ab\ | + | This looks a lot like Law of Cosines, which is <math>c^2=a^2+b^2-2ab\cos{c}</math>. |
− | <cmath>c^2=a^2+b^2-ab=a^2+b^2-2ab\ | + | <cmath>c^2=a^2+b^2-ab=a^2+b^2-2ab\cos{c}</cmath> |
− | <cmath>ab=2ab\ | + | <cmath>ab=2ab\cos{c}</cmath> |
− | <cmath>\frac{1}{2}=\ | + | <cmath>\frac{1}{2}=\cos{c}</cmath> |
− | <math>\ | + | <math>\cos{c}</math> is <math>\frac{1}{2}</math>, so the angle opposite side <math>c</math> is <math>\boxed{60^\circ}</math>. |
-aopspandy | -aopspandy |
Latest revision as of 19:20, 18 June 2021
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
.
Solution 2
This looks a lot like Law of Cosines, which is
.
is
, so the angle opposite side
is
.
-aopspandy