Difference between revisions of "2010 AMC 12B Problems/Problem 13"
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We note that <math>-1</math> <math>\le</math> <math>\sin x</math> <math>\le</math> <math>1</math> and <math>-1</math> <math>\le</math> <math>\cos x</math> <math>\le</math> <math>1</math>. | We note that <math>-1</math> <math>\le</math> <math>\sin x</math> <math>\le</math> <math>1</math> and <math>-1</math> <math>\le</math> <math>\cos x</math> <math>\le</math> <math>1</math>. | ||
Therefore, the only way to satisfy this equation is if both <math>\cos(2A-B)=1</math> and <math>\sin(A+B)=1</math>, since if either one of these is less than 1, the other one would have to be greater than 1, which contradicts our previous statement. | Therefore, the only way to satisfy this equation is if both <math>\cos(2A-B)=1</math> and <math>\sin(A+B)=1</math>, since if either one of these is less than 1, the other one would have to be greater than 1, which contradicts our previous statement. | ||
− | From this we can easily conclude that <math>2A-B=0^{\circ}</math> and <math>A+B=90^{\circ}</math> and solving this system gives us <math>A=30^{\circ}</math> and <math>B=60^{\circ}</math>. | + | From this we can easily conclude that <math>2A-B=0^{\circ}</math> and <math>A+B=90^{\circ}</math> and solving this system gives us <math>A=30^{\circ}</math> and <math>B=60^{\circ}</math>. We can easily see that <math>\triangle ABC</math> is a <math>30^{\circ}-60^{\circ}-90^{\circ}</math> triangle, and solving for the sides gives us <math>BC=2</math> <math>\Longrightarrow</math> <math>(C)</math> |
Revision as of 23:06, 6 April 2010
Problem
In ,
and
. What is
?
Solution
We note that
and
.
Therefore, the only way to satisfy this equation is if both
and
, since if either one of these is less than 1, the other one would have to be greater than 1, which contradicts our previous statement.
From this we can easily conclude that
and
and solving this system gives us
and
. We can easily see that
is a
triangle, and solving for the sides gives us