Difference between revisions of "2023 AMC 12A Problems/Problem 15"
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==Solution 1== | ==Solution 1== | ||
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By "unfolding" line APQRS into a straight line, we get a right triangle ABS. | By "unfolding" line APQRS into a straight line, we get a right triangle ABS. | ||
| − | < | + | |
| − | < | + | <math>cos(\theta)=\frac{120}{100}</math> |
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| + | <math>\theta=\boxed{\textbf{(A) } cos^-1(\frac{5}{6})}</math> | ||
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| + | ==See also== | ||
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| + | {{AMC12 box|year=2023|ab=A|num-b=14|num-a=16}} | ||
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| + | {{MAA Notice}} | ||
Revision as of 22:46, 9 November 2023
Question
[uh someone insert diagram]
Solution 1
By "unfolding" line APQRS into a straight line, we get a right triangle ABS.
See also
| 2023 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 14 |
Followed by Problem 16 |
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
