Question

抛物线 $y = x^2 + 2x - 15$ 与 $x$ -轴交于点 $A$ 、 $C$ ，与 $y$ -轴交于点 $B$ 。求 $\tan(\angle ABC)$ 。

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 $\tan(\angle ABC)?$

$\dfrac17$

$\dfrac14$

$\dfrac37$

$\dfrac12$

$\dfrac47$

答案：E

知识点：坐标几何向量三角学

难度评级：1570

解答：

因式分解得 $x^2 + 2x - 15 = (x+5)(x-3)$ 所以 $A = (-5, 0)$ ， $C = (3, 0)$ 且 $y$ -截距为 $B = (0, -15)$ 。

于是 $\vec{BA} = (-5, 15)$ ， $\vec{BC} = (3, 15)$ 利用叉积和点积， $\begin{gathered} \tan(\angle ABC) = \scriptsize \dfrac{|(-5)(15) - (15)(3)|}{(-5)(3) + (15)(15)} \\ = \dfrac{120}{210} \\ = \dfrac47. \end{gathered}$

所以正确答案是 E。

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, $\begin{gathered} \tan(\angle ABC) = \scriptsize \dfrac{|(-5)(15) - (15)(3)|}{(-5)(3) + (15)(15)} \\ = \dfrac{120}{210} \\ = \dfrac47. \end{gathered}$

Thus, the correct answer is E.

2022 AMC 12B 第 14 题

答案：E

难度评级：1570

解答：

其他年份的第 14 题