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

Revision as of 19:59, 19 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

$y=x^2+2x-15$ 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 2

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 3

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 $(-3-5)/(1+(-3*5))$ which is $\boxed{\textbf{(E)}\ \frac{4}{7}}.$

~mathboy100 (minor LaTeX edits)

@@ Line 29: / Line 29: @@
 ==Solution 3==
-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*5))</math> which is <math>\boxed{\textbf{(E)}\ \frac{4}{7}}.</math>
+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*5))</math> which is <math>\boxed{\textbf{(E)}\ \frac{4}{7}}.</math>
 ~mathboy100 (minor LaTeX edits)

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