Difference between revisions of "2021 Fall AMC 12A Problems/Problem 19"
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− | Therefore, the answer is <math>\boxed{\textbf{(B) }13},</math> as <math>\sin 11^\circ - \sin 10^\circ</math> is closest to <math>0.</math> | + | Therefore, the answer is <math>\boxed{\textbf{(B) }13},</math> as <math>\sin 11^\circ - \sin 10^\circ</math> is the closest to <math>0.</math> |
~Steven Chen (www.professorchenedu.com) | ~Steven Chen (www.professorchenedu.com) |
Revision as of 02:07, 26 November 2021
Contents
Problem
Let be the least real number greater than
such that
, where the arguments are in degrees. What is
rounded up to the closest integer?
Solution 1
The smallest to make
would require
, but since
needs to be greater than
, these solutions are not valid.
The next smallest would require
, or
.
After a bit of guessing and checking, we find that , and
, so the solution lies between
and
, making our answer
Note: One can also solve the quadratic and estimate the radical.
~kingofpineapplz
Solution 2
For choice we have
For choice
we have
For choice
we have
For choice
we have
For choice
we have
Therefore, the answer is
as
is the closest to
~Steven Chen (www.professorchenedu.com)
See Also
2021 Fall AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
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.