Difference between revisions of "2024 AMC 12A Problems/Problem 10"
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==Solution 2: Trial and Error == | ==Solution 2: Trial and Error == | ||
Another approach to solving this problem is trial and error, comparing the sine of the answer choices with <math>\sin\beta = \frac{7}{25}</math>. Starting with the easiest sine to compute from the answer choices (option choice D). We get: | Another approach to solving this problem is trial and error, comparing the sine of the answer choices with <math>\sin\beta = \frac{7}{25}</math>. Starting with the easiest sine to compute from the answer choices (option choice D). We get: | ||
| − | <cmath>\sin{(\frac{\alpha}{2})} = \sqrt{\frac{1 - \cos\alpha}{2}}</cmath> | + | <cmath>\sin{(\frac{\alpha}{2})} = \sqrt{\frac{1 - \cos{\alpha}}{2}}</cmath> |
<cmath>= \sqrt{\frac{1 - \frac{4}{5}}{2}}</cmath> | <cmath>= \sqrt{\frac{1 - \frac{4}{5}}{2}}</cmath> | ||
<cmath>= \sqrt{\frac{1}{10}}</cmath> | <cmath>= \sqrt{\frac{1}{10}}</cmath> | ||
| Line 23: | Line 23: | ||
The next easiest sine to compute is option choice C. | The next easiest sine to compute is option choice C. | ||
| − | <cmath>\sin(\frac{\pi}{2} - 2\alpha) = \sin{(\frac{\pi}{2})}\cos{(2\alpha)}</cmath> | + | <cmath>\sin{(\frac{\pi}{2} - 2\alpha)} = \sin{(\frac{\pi}{2})}\cos{(2\alpha)}</cmath> |
<cmath>=\cos{2\alpha}</cmath> | <cmath>=\cos{2\alpha}</cmath> | ||
<cmath>=\cos^2{\alpha} - \sin^2{\alpha}</cmath> | <cmath>=\cos^2{\alpha} - \sin^2{\alpha}</cmath> | ||
Revision as of 18:59, 8 November 2024
Problem
Let 
be the radian measure of the smallest angle in a 
right triangle. Let 
be the radian measure of the smallest angle in a 
right triangle. In terms of , what is ?
Solution 1
From question,
![\[tan\alpha=\frac{3}{4}, \space tan\beta=\frac{7}{24}\]](http://latex.artofproblemsolving.com/3/7/c/37c0129c887e266b41bc705db74322106c6d22a3.png)
![\[tan(\alpha+\beta)= \frac{tan\alpha+tan\beta}{1-tan\alpha tan\beta}\]](http://latex.artofproblemsolving.com/a/d/e/adea8f2d55bde236dfb35c4e1e2794ba2ae0f6da.png)
![\[tan(\alpha+\beta)= \frac{\frac{3}{4}+\frac{7}{24}}{1-\frac{3}{4} \cdot \frac{7}{24}}\]](http://latex.artofproblemsolving.com/2/0/a/20ac87bb8453be75cac2dc1ff053a7efdf992dea.png)
![\[tan(\alpha+\beta)=\frac{4}{3}\]](http://latex.artofproblemsolving.com/c/7/9/c790151177f48defd8e4fbae29f9a74b13c7f243.png)
![\[\alpha+\beta=tan^{-1}(\frac{4}{3})\]](http://latex.artofproblemsolving.com/b/5/c/b5c523e84b0a356d1c249bbbdac2525a5900a102.png)
![\[\alpha+\beta=\frac{\pi}{2}-\alpha\]](http://latex.artofproblemsolving.com/2/f/3/2f3db36a2888ec0c1261832bb95f80f5368e076f.png)
$$ (Error compiling LaTeX. Unknown error_msg)\beta=\fbox{(C) 
}$$ (Error compiling LaTeX. Unknown error_msg)
~lptoggled
Solution 2: Trial and Error
Another approach to solving this problem is trial and error, comparing the sine of the answer choices with 
. Starting with the easiest sine to compute from the answer choices (option choice D). We get:
![\[\sin{(\frac{\alpha}{2})} = \sqrt{\frac{1 - \cos{\alpha}}{2}}\]](http://latex.artofproblemsolving.com/e/c/8/ec8cc802e754f6d7cdb7bdfb5c6a46bda95f0163.png)
![\[= \sqrt{\frac{1 - \frac{4}{5}}{2}}\]](http://latex.artofproblemsolving.com/6/2/8/6285d4f10dd2b2d463b07b6b5b31142e129017d8.png)
![\[= \sqrt{\frac{1}{10}}\]](http://latex.artofproblemsolving.com/2/a/8/2a89c614d76def405701996466ba610d43b74c45.png)
![\[\neq \frac{7}{25}\]](http://latex.artofproblemsolving.com/b/4/c/b4c6a0bdf975759c073a6e0cdab4bf4944591d7b.png)
The next easiest sine to compute is option choice C.
![\[\sin{(\frac{\pi}{2} - 2\alpha)} = \sin{(\frac{\pi}{2})}\cos{(2\alpha)}\]](http://latex.artofproblemsolving.com/f/8/e/f8ed04c29a1c1d24c1861052a433547790a993d7.png)
![\[=\cos{2\alpha}\]](http://latex.artofproblemsolving.com/6/3/b/63bb2aee4d27bd6bd2800b96aa09be1e4721e090.png)
![\[=\cos^2{\alpha} - \sin^2{\alpha}\]](http://latex.artofproblemsolving.com/7/4/8/748231d1fb7f1cc4112820b455a9e8aa5535997f.png)
![\[=\frac{16}{25} - \frac{9}{25}\]](http://latex.artofproblemsolving.com/e/6/6/e6674d0d8845bb7495ae28e8efb94fd895e7c16f.png)
![\[=\frac{7}{25}\]](http://latex.artofproblemsolving.com/0/8/6/086d6daa24bb018a7d19505a1a5daacf9f7080c6.png)
Since 
is equal to 
, option choice C is the correct answer. ~amshah
See also
| 2024 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 9 |
Followed by Problem 11 |
| 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. 
