Difference between revisions of "2024 AMC 12A Problems/Problem 10"
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<math>\textbf{(A) }\frac{\alpha}{3}\qquad \textbf{(B) }\alpha - \frac{\pi}{8}\qquad \textbf{(C) }\frac{\pi}{2} - 2\alpha \qquad \textbf{(D) }\frac{\alpha}{2}\qquad \textbf{(E) }\pi - 4\alpha\qquad</math> | <math>\textbf{(A) }\frac{\alpha}{3}\qquad \textbf{(B) }\alpha - \frac{\pi}{8}\qquad \textbf{(C) }\frac{\pi}{2} - 2\alpha \qquad \textbf{(D) }\frac{\alpha}{2}\qquad \textbf{(E) }\pi - 4\alpha\qquad</math> | ||
==Solution 1== | ==Solution 1== | ||
| − | From question, | + | From the question, |
| − | <cmath>tan\alpha=\frac{3}{4}, \space tan\beta=\frac{7}{24}</cmath> | + | <cmath>\tan\alpha=\frac{3}{4}, \space \tan\beta=\frac{7}{24}</cmath> |
| − | <cmath>tan(\alpha+\beta)= \frac{tan\alpha+tan\beta}{1-tan\alpha tan\beta}</cmath> | + | <cmath>\tan(\alpha+\beta)= \frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}</cmath> |
| − | <cmath>tan(\alpha+\beta)= \frac{\frac{3}{4}+\frac{7}{24}}{1-\frac{3}{4} \cdot \frac{7}{24}}</cmath> | + | <cmath>\tan(\alpha+\beta)= \frac{\frac{3}{4}+\frac{7}{24}}{1-\frac{3}{4} \cdot \frac{7}{24}}</cmath> |
| − | <cmath>tan(\alpha+\beta)=\frac{4}{3}</cmath> | + | <cmath>\tan(\alpha+\beta)=\frac{4}{3}</cmath> |
| − | <cmath>\alpha+\beta=tan^{-1}(\frac{4}{3})</cmath> | + | <cmath>\alpha+\beta=\tan^{-1}(\frac{4}{3})</cmath> |
<cmath>\alpha+\beta=\frac{\pi}{2}-\alpha</cmath> | <cmath>\alpha+\beta=\frac{\pi}{2}-\alpha</cmath> | ||
<cmath>\beta=\boxed{(C) \frac{\pi}{2} -2\alpha}</cmath> | <cmath>\beta=\boxed{(C) \frac{\pi}{2} -2\alpha}</cmath> | ||
~lptoggled | ~lptoggled | ||
| − | |||
==Solution 2: Trial and Error == | ==Solution 2: Trial and Error == | ||
Revision as of 19:57, 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 the question,
![\[\tan\alpha=\frac{3}{4}, \space \tan\beta=\frac{7}{24}\]](http://latex.artofproblemsolving.com/9/d/8/9d8077a0df7a459f13e18cd1f65892a48ba03695.png)
![\[\tan(\alpha+\beta)= \frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}\]](http://latex.artofproblemsolving.com/5/2/f/52fa38e4d1f401474de46e9414366c6f7054c682.png)
![\[\tan(\alpha+\beta)= \frac{\frac{3}{4}+\frac{7}{24}}{1-\frac{3}{4} \cdot \frac{7}{24}}\]](http://latex.artofproblemsolving.com/1/9/8/198fa0e3f8bd04b8b106ff46bf14c805f286e616.png)
![\[\tan(\alpha+\beta)=\frac{4}{3}\]](http://latex.artofproblemsolving.com/3/f/0/3f0ba134eb4fad67f748f1cbe46a5b075d09d349.png)
![\[\alpha+\beta=\tan^{-1}(\frac{4}{3})\]](http://latex.artofproblemsolving.com/9/3/8/938aa86a2248c5bddf44a6f7de80b2b5e56b8840.png)
![\[\alpha+\beta=\frac{\pi}{2}-\alpha\]](http://latex.artofproblemsolving.com/2/f/3/2f3db36a2888ec0c1261832bb95f80f5368e076f.png)
![\[\beta=\boxed{(C) \frac{\pi}{2} -2\alpha}\]](http://latex.artofproblemsolving.com/f/4/0/f40a0ee871c23fee2f00797d7ea8a819b4c2ccb9.png)
~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. 
