Difference between revisions of "2022 AMC 12B Problems/Problem 14"

Revision as of 21:09, 20 November 2022

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)

@@ Line 9: / Line 9: @@
 \textbf{(E)}\ \frac{4}{7} \qquad</math>
-== Solution 1==
+== Solution 1 (Dot Product) ==
+First, find <math>A=(-5,0)</math>, <math>B=(0,-15)</math>, and <math>C=(3,0)</math>. Create vectors <math>\overrightarrow{BA}</math> and <math>\overrightarrow{BC}.</math> These can be reduced to <math>\langle -1, 3 \rangle</math> and <math>\langle 1, 5 \rangle</math>, respectively. Then, we can use the dot product to calculate the cosine of the angle (where <math>\theta=\angle ABC</math>) between them:
+<cmath>
+\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*}
+</cmath>
+Thus, <cmath>\tan(\angle ABC) = \sqrt{\frac{65}{49}-1}= \boxed{\textbf{(E)}\ \frac{4}{7}}.</cmath>
+~Indiiiigo
+== Solution 2==
 <math>y=x^2+2x-15</math> intersects the <math>x</math>-axis at points <math>(-5, 0)</math> and <math>(3, 0)</math>. Without loss of generality, let these points be <math>A</math> and <math>C</math> respectively. Also, the graph intersects the y-axis at point <math>B = (0, -15)</math>.
@@ Line 21: / Line 36: @@
 Alternatively, we can use the [[Pythagorean Theorem]] to find that <math>AB = 5 \sqrt{10}</math> and <math>BC = 3 \sqrt{26}</math> and then use the <math>A = \frac12 ab \sin \angle C</math> [[Area of a Triangle|area formula for a triangle]] and the [[Law of Cosines]] to find <math>\tan(\angle ABC)</math>.
-==Solution 2==
+==Solution 3==
 Like above, we set <math>A</math> to <math>(-5,0)</math>, <math>B</math> to <math>(0, -15)</math>, and <math>C</math> to <math>(3,0)</math>, then finding via the Pythagorean Theorem that <math>AB = 5 \sqrt{10}</math> and <math>BC = 3 \sqrt{26}</math>. Using the Law of Cosines, we see that <cmath>\cos(\angle ABC) = \frac{AB^2 + BC^2 - AC^2}{2 AB BC} = \frac{250 + 234 - 64}{15 \sqrt{260}} = \frac{7}{\sqrt{65}}.</cmath> Then, we use the identity <math>\tan^2(x) = \sec^2(x) - 1</math> to get <cmath>\tan(\angle ABC) = \sqrt{\frac{65}{49} - 1} = \boxed{\textbf{(E)}\ \frac{4}{7}}.</cmath>
@@ Line 27: / Line 42: @@
 ~ jamesl123456
-==Solution 3==
+==Solution 4==
 We can reflect the figure, but still have the same angle. This problem is the same as having points <math>D(0,0)</math>, <math>E(3,15)</math>, and <math>F(-5,15)</math>, where we're solving for angle FED. We can use the formula for <math>\tan(a-b)</math> to solve now where <math>a</math> is the <math>x</math>-axis to angle <math>F</math> and <math>b</math> is the <math>x</math>-axis to angle <math>E</math>. <math>\tan(a) = \textrm{slope of line }DF = -3</math> and <math>\tan(B) = \textrm{slope of line }DE = 5</math>. Plugging these values into the <math>\tan(a-b)</math> formula, we get <math>(-3-5)/(1+(-3\cdot 5))</math> which is <math>\boxed{\textbf{(E)}\ \frac{4}{7}}.</math>

2022 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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