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== | ||
+ | From question, | ||
+ | <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{\frac{3}{4}+\frac{7}{24}}{1-\frac{3}{4} \cdot \frac{7}{24}}</cmath> | ||
+ | <cmath>tan(\alpha+\beta)=\frac{4}{3}</cmath> | ||
+ | <cmath>\alpha+\beta=tan^{-1}(\frac{4}{3})</cmath> | ||
+ | <cmath>\alpha+\beta=\frac{\pi}{2}-\alpha</cmath> | ||
+ | <math></math>\beta=\fbox{(C) <math>\frac{\pi}{2} -2\alpha</math>}<math></math> | ||
==See also== | ==See also== | ||
{{AMC12 box|year=2024|ab=A|num-b=9|num-a=11}} | {{AMC12 box|year=2024|ab=A|num-b=9|num-a=11}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 18:08, 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, $$ (Error compiling LaTeX. Unknown error_msg)\beta=\fbox{(C) }$$ (Error compiling LaTeX. Unknown error_msg)
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 |
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