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2022 AMC 12B Problem 14

Problem 14 of 25IntermediateGeometry

The graph of y=x2+2x−15y = x^2 + 2x - 15 intersects the xx-axis at points AA and CC and the yy-axis at point B.B. What is tan⁡(∠ABC)?\tan(\angle ABC)?

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Solution

Factoring, x2+2x−15=(x+5)(x−3),x^2 + 2x - 15 = (x+5)(x-3), so A=(−5,0)A = (-5, 0) and C=(3,0),C = (3, 0), and the yy-intercept is B=(0,−15).B = (0, -15). Then BA⃗=(−5,15)\vec{BA} = (-5, 15) and BC⃗=(3,15).\vec{BC} = (3, 15). Using the cross and dot products, tan⁡(∠ABC)=∣(−5)(15)−(15)(3)∣(−5)(3)+(15)(15)=120210=47. \begin{gathered} \tan(\angle ABC) = \scriptsize \dfrac{|(-5)(15) - (15)(3)|}{(-5)(3) + (15)(15)} \\ = \dfrac{120}{210} \\ = \dfrac47. \end{gathered} Thus, the correct answer is E.
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Tagged: coordinate geometry · vector · trigonometry

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