1981 AHSME Problems/Problem 21
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 $c^2=a^2+b^2-2ab\cosc$ (Error compiling LaTeX. Unknown error_msg)
\[c^2=a^2+b^2-ab=a^2+b^2-2ab\cosc\] (Error compiling LaTeX. Unknown error_msg)
\[ab=2ab\cosc\] (Error compiling LaTeX. Unknown error_msg)
\[\frac{1}{2}=\cosc\] (Error compiling LaTeX. Unknown error_msg)
$\cosc$ (Error compiling LaTeX. Unknown error_msg) is c
\boxed{60^\circ}$
-aopspandy