AMC12 2022 B

AMC12 2022 B · Q14

AMC12 2022 B · Q14. It mainly tests Coordinate geometry, Trigonometry (basic).

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 $\tan(\angle ABC)$?

$y=x^2+2x-15$ 的图像与 $x$ 轴相交于点 $A$ 和 $C$，与 $y$ 轴相交于点 $B$。$\tan(\angle ABC)$ 是多少？

(A) \ \frac{1}{7} \ \frac{1}{7}

(B) \ \frac{1}{4} \ \frac{1}{4}

(D) \ \frac{1}{2} \ \frac{1}{2}

(E) \ \frac{4}{7} \qquad \ \frac{4}{7} \qquad

Answer

Correct choice: (E)

正确答案：（E）

Solution

First, find $A=(-5,0)$, $B=(0,-15)$, and $C=(3,0)$. Create vectors $\overrightarrow{BA}$ and $\overrightarrow{BC}.$ These can be reduced to $\langle -1, 3 \rangle$ and $\langle 1, 5 \rangle$, respectively. Then, we can use the dot product to calculate the cosine of the angle (where $\theta=\angle ABC$) between them: \begin{align*} \langle -1, 3 \rangle \cdot \langle 1, 5 \rangle = 15-1 &= \sqrt{10}\sqrt{26}\cos(\theta),\\ \implies \cos (\theta) &= \frac{7}{\sqrt{65}}. \end{align*} Thus, \[\tan(\angle ABC) = \sqrt{\frac{65}{49}-1}= \boxed{\textbf{(E)}\ \frac{4}{7}}.\]

首先，求得 $A=(-5,0)$，$B=(0,-15)$，$C=(3,0)$。创建向量 $\overrightarrow{BA}$ 和 $\overrightarrow{BC}$。它们可以简化为 $\langle -1, 3 \rangle$ 和 $\langle 1, 5 \rangle$。然后，使用点积计算它们之间夹角（其中 $\theta=\angle ABC$）的余弦： \begin{align*} \langle -1, 3 \rangle \cdot \langle 1, 5 \rangle = 15-1 &= \sqrt{10}\sqrt{26}\cos(\theta),\\ \implies \cos (\theta) &= \frac{7}{\sqrt{65}}. \end{align*} 因此，\[\tan(\angle ABC) = \sqrt{\frac{65}{49}-1}= \boxed{\textbf{(E)}\ \frac{4}{7}}.\]

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