2022 AMC 12B Problems/Problem 14

Problem

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 $\tan(\angle ABC)$ ?

$\textbf{(A)}\ \frac{1}{7} \qquad \textbf{(B)}\ \frac{1}{4} \qquad \textbf{(C)}\ \frac{3}{7} \qquad \textbf{(D)}\ \frac{1}{2} \qquad \textbf{(E)}\ \frac{4}{7} \qquad$

Solution 1 (Dot Product)

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, $\tan(\angle ABC) = \sqrt{\frac{65}{49}-1}= \boxed{\textbf{(E)}\ \frac{4}{7}}.$

~Indiiiigo

Solution 2

$y=x^2+2x-15$ intersects the $x$ -axis at points $(-5, 0)$ and $(3, 0)$ . Without loss of generality, let these points be $A$ and $C$ respectively. Also, the graph intersects the y-axis at point $B = (0, -15)$ .

Let point $O$ denote the origin $(0, 0)$ . Note that triangles $AOB$ and $BOC$ are right.

We have

$\tan(\angle ABC) = \tan(\angle ABO + \angle OBC) = \frac{\tan(\angle ABO) + \tan(\angle OBC)}{1 - \tan(\angle ABO) \cdot \tan(\angle OBC)} = \frac{\frac15 + \frac13}{1 - \frac1{15}} = \boxed{\textbf{(E)}\ \frac{4}{7}}.$

Alternatively, we can use the Pythagorean Theorem to find that $AB = 5 \sqrt{10}$ and $BC = 3 \sqrt{26}$ and then use the $A = \frac12 ab \sin \angle C$ area formula for a triangle and the Law of Cosines to find $\tan(\angle ABC)$ .

Solution 3

Like above, we set $A$ to $(-5,0)$ , $B$ to $(0, -15)$ , and $C$ to $(3,0)$ , then finding via the Pythagorean Theorem that $AB = 5 \sqrt{10}$ and $BC = 3 \sqrt{26}$ . Using the Law of Cosines, we see that $\cos(\angle ABC) = \frac{AB^2 + BC^2 - AC^2}{2 AB BC} = \frac{250 + 234 - 64}{15 \sqrt{260}} = \frac{7}{\sqrt{65}}.$ Then, we use the identity $\tan^2(x) = \sec^2(x) - 1$ to get $\tan(\angle ABC) = \sqrt{\frac{65}{49} - 1} = \boxed{\textbf{(E)}\ \frac{4}{7}}.$

~ jamesl123456

Solution 4

We can reflect the figure, but still have the same angle. This problem is the same as having points $D(0,0)$ , $E(3,15)$ , and $F(-5,15)$ , where we're solving for angle FED. We can use the formula for $\tan(a-b)$ to solve now where $a$ is the $x$ -axis to angle $F$ and $b$ is the $x$ -axis to angle $E$ . $\tan(a) = \textrm{slope of line }DF = -3$ and $\tan(B) = \textrm{slope of line }DE = 5$ . Plugging these values into the $\tan(a-b)$ formula, we get $(-3-5)/(1+(-3\cdot 5))$ which is $\boxed{\textbf{(E)}\ \frac{4}{7}}.$

~mathboy100 (minor LaTeX edits)

Solution 5

We use the identity $[ABC]=\frac{1}{2}ab\sin{C}.$

Note that $\triangle ABC$ has side-lengths $AB=5\sqrt{10}$ and $BC=3\sqrt{26}$ from Pythagorean theorem, with the area being $\frac12\cdot8\cdot15.$

We equate the areas together to get: $\frac12\cdot8\cdot15=\frac12\cdot5\sqrt{10}\cdot3\sqrt{26}\cdot\sin{B},$ from which $\sin{B}=\frac{8}{\sqrt{260}}.$

From Pythagorean identity, $\cos{B}=\frac{14}{\sqrt{260}}.$

Then we use $\tan{B}=\frac{\sin{B}}{\cos{B}}$ , to obtain $\tan{B}=\frac{8}{14}=\boxed{\textbf{(E)}\ \frac{4}{7}}.$

- SAHANWIJETUNGA

2022 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

Page

Toolbox

Search