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2022 AMC 12B Problems/Problem 14

Revision as of 15:45, 18 November 2022 by Jamesl123456 (talk | contribs)

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 that $AB = 5 \sqrt{10}$ and $BC = 3 \sqrt{26}$. Using the Law of Cosines, we can then find 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 \[tan(x) = \sqrt{\frac{65}{49} - 1} = \boxed{\textbf{(E)}\ \frac{4}{7}}.\]

~ jamesl123456

See Also

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

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