Difference between revisions of "2010 AMC 12B Problems/Problem 13"
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− | We | + | 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, | + | Therefore, there is no other way to satisfy this equation other than making both <math>\cos(2A-B)=1</math> and <math>\sin(A+B)=1</math>, since any other way would cause one of these values to become greater than 1, which contradicts our previous statement. |
− | From this we easily conclude that <math>2A-B=0</math> and <math>A+B=90</math> and solving this system gives us <math>A=30</math> and <math>B=60</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>. It is clear that <math>\triangle ABC</math> is a <math>30^{\circ},60^{\circ},90^{\circ}</math> triangle with <math>BC=2</math> <math>\Longrightarrow</math> <math>(C)</math>. |
+ | |||
+ | == See also == | ||
+ | {{AMC12 box|year=2010|num-b=12|num-a=14|ab=B}} | ||
+ | |||
+ | [[Category:Introductory Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 23:31, 17 January 2020
Problem
In , and . What is ?
Solution
We note that and . Therefore, there is no other way to satisfy this equation other than making both and , since any other way would cause one of these values to become greater than 1, which contradicts our previous statement. From this we can easily conclude that and and solving this system gives us and . It is clear that is a triangle with .
See also
2010 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 12 |
Followed by Problem 14 |
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 | |
All AMC 12 Problems and Solutions |
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