2024 AMC 12A Problem 10TrigonometryMediumLet 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−2αSolution 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−2αFinal Answer(C) π2−2α\frac{\pi}{2}-2\alpha2π−2αRelated Problems (11)AMC 12B 2024 (Problem 21)AIME II 2025 (Problem 9)AMC 12A 2024 (Problem 8)AMC 10A 2025 (Problem 15)AMC 12B 2025 (Problem 16)AIME II 2026 (Problem 8)2024 AMC 12A Problem 23AMC 10B 2024 (Problem 10)AMC 10B/12B 2024 (Problem 11/7)AMC 12B 2024 (Problem 19)AMC 12B 2024 (Problem 22)View all related problems →
