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

(→‎Solution: I really hate the tangent addition formula, but it's admittedly the undoubtly best sol here)
(→‎Solution: students who haven't taken PC may enjoy the second (brutal calculation) method)
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Let point <math>O = (0, 0)</math>. Note that triangles <math>AOB</math> and <math>BOC</math> are right.  
 
Let point <math>O = (0, 0)</math>. Note that triangles <math>AOB</math> and <math>BOC</math> are right.  
  
−
<cmath>\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}}.
+
<cmath>\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}}.</cmath>
  
−
Alternatively, we can use the [[Pythagorean Theorem]] to find that using the sin and cos angle addition formulas:
+
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 formula for a triangle and the [[Law of Cosines]] to find <math>\tan(\angle ABC)</math>.
−
 
 
−
</cmath>\tan(\angle ABC) = \frac{\sin (\angle ABC)}{\cos (\angle ABC)} = \frac{\sin(\angle ABO + \angle OBC)}{\cos(\angle ABO + \angle OBC)} = \frac{\sin (\angle ABO) \cdot \cos (\angle OBC) + \cos (\angle ABO) \cdot \sin(\angle OBC)}{<math></math>
 
−
 
 
−
Alternatively, the
 
  
 
== See Also ==
 
== See Also ==
 
{{AMC12 box|year=2022|ab=B|num-b=13|num-a=15}}
 
{{AMC12 box|year=2022|ab=B|num-b=13|num-a=15}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 19:10, 17 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

$y=x^2+2x-15$ 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 $O = (0, 0)$. Note that triangles $AOB$ and $BOC$ are right.

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

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 $\tan(\angle ABC)$.

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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