2024 AMC 12A Problem 10Let α\alphaα be the radian measure of the smallest angle in a 3−4−53-4-53−4−5 right triangle. Let β\betaβ be the radian measure of the smallest angle in a 7−24−257-24-257−24−25 right triangle. In terms of α\alphaα, what is β\betaβ?(A) α3\text{(A)}\;\frac{\alpha}{3}(A)3α(B) α−π8\text{(B)}\;\alpha-\frac{\pi}{8}(B)α−8π(C) π2−2α\text{(C)}\;\frac{\pi}{2}-2\alpha(C)2π−2α(D) α2\text{(D)}\;\frac{\alpha}{2}(D)2α(E) π−4α\text{(E)}\;\pi-4\alpha(E)π−4αSolution 1Related TopicsCoreSimilar TrianglesCheck AnswerYour answer:ABCDECheckHints (4)Hint 1Create a triangle by sticking two triangles similar or (itself)to the triangles we have.Hint 2Hint 3Multiply the sides of smaller triangle byHint 4BC=7+18=25=ABBC=7+18=25=ABBC=7+18=25=AB⇒∠BAC=∠BCA\Rightarrow \angle BAC=\angle BCA⇒∠BAC=∠BCAα+β=π2−α\alpha+\beta=\frac{\pi}{2}-\alphaα+β=2π−αβ=π2−2α\beta=\frac{\pi}{2}-2\alphaβ=2π−2αRelated Problems (1)AMC 10A 2025 (Problem 15)Solution 2Related TopicsCoreTrigonometric IdentitiesTrigonometric FoundationsMinorTrigonometric TransformationsHints (4)Hint 1cosα=45cosβ=2425\cos\alpha=\frac{4}{5}\qquad\cos\beta=\frac{24}{25}cosα=54cosβ=2524Hint 2cos(π2−2α)=sin2α\cos\left(\frac{\pi}{2}-2\alpha\right)=\sin2\alphacos(2π−2α)=sin2αHint 3sin2α=2sinαcosα=2(35)(45)=2425=cosβ\sin2\alpha=2\sin\alpha\cos\alpha=2\left(\frac{3}{5}\right)\left(\frac{4}{5}\right)=\frac{24}{25}=\cos\betasin2α=2sinαcosα=2(53)(54)=2524=cosβHint 4sin2α=sin(π2−β)\sin2\alpha=\sin\left(\frac{\pi}{2}-\beta\right)sin2α=sin(2π−β)0<α<π4⇒0<2α<π20<\alpha<\frac{\pi}{4}\quad\Rightarrow\quad0<2\alpha<\frac{\pi}{2}0<α<4π⇒0<2α<2π0<β<π2⇒0<π2−β<π20<\beta<\frac{\pi}{2}\quad\Rightarrow\quad0<\frac{\pi}{2}-\beta<\frac{\pi}{2}0<β<2π⇒0<2π−β<2π⇒2α=π2−β⇒β=π2−2α\Rightarrow 2\alpha=\frac{\pi}{2}-\beta\quad\Rightarrow\quad\beta=\frac{\pi}{2}-2\alpha⇒2α=2π−β⇒β=2π−2αRelated Problems (5)AIME II 2025 (Problem 9)AMC 12A 2024 (Problem 8)AMC 12B 2025 (Problem 16)AIME II 2026 (Problem 8)2024 AMC 12A Problem 23Final Answer(C) π2−2α\frac{\pi}{2}-2\alpha2π−2α
