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> | ||
+ | |||
+ | ==Solution== | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | <cmath>(a+b+c)(a+b-c)=3ab</cmath> | ||
+ | <cmath>a^2+2ab+b^2-c^2=3ab</cmath> | ||
+ | <cmath>a^2+b^2-c^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\cos{c}</math>. | ||
+ | <cmath>c^2=a^2+b^2-ab=a^2+b^2-2ab\cos{c}</cmath> | ||
+ | <cmath>ab=2ab\cos{c}</cmath> | ||
+ | <cmath>\frac{1}{2}=\cos{c}</cmath> | ||
+ | <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 |
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