2022 AMC 12B Problem 14

Below is the professionally curated solution for Problem 14 of the 2022 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2022 AMC 12B solutions, or check the answer key.

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Concepts:coordinate geometry vector trigonometry

Difficulty rating: 1570

14.

The graph of $y = x^2 + 2x - 15$ intersects the $x$ -axis at points $A$ and $C$ and the $y$ -axis at point $B.$ What is $\tan(\angle ABC)?$

$\dfrac17$

$\dfrac14$

$\dfrac37$

$\dfrac12$

$\dfrac47$

Solution:

Factoring, $x^2 + 2x - 15 = (x+5)(x-3),$ so $A = (-5, 0)$ and $C = (3, 0),$ and the $y$ -intercept is $B = (0, -15).$

Then $\vec{BA} = (-5, 15)$ and $\vec{BC} = (3, 15).$ Using the cross and dot products, $\tan(\angle ABC) = \dfrac{|(-5)(15) - (15)(3)|}{(-5)(3) + (15)(15)} = \dfrac{120}{210} = \dfrac47.$

Thus, the correct answer is E.

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