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

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~ jamesl123456
 
~ jamesl123456
  
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Alternatively, we could observe that <math>\sin(\angle ABC)=\sqrt{1-\cos^2(\angle ABC)}=\sqrt{1-\dfrac7{\sqrt{65}}}=\sqrt{{16}{65}}=\dfrac4{\sqrt{65}}</math>, so <math>\tan(\angle ABC)=\dfrac{\sin(\angle ABC)}{\cos(\angle ABC)}=\boxed{\texbf{(E) }\dfrac47}</math>.  
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Alternatively, we could observe that <math>\sin(\angle ABC)=\sqrt{1-\cos^2(\angle ABC)}=\sqrt{1-\dfrac7{\sqrt{65}}}=\sqrt{{16}{65}}=\dfrac4{\sqrt{65}}</math>, so <math>\tan(\angle ABC)=\dfrac{\sin(\angle ABC)}{\cos(\angle ABC)}=\boxed{\textbf{(E) }\dfrac47}</math>.  
  
 
~Technodoggo
 
~Technodoggo

Revision as of 23:08, 1 November 2024

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 $\tan(\angle ABC)$?

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

Diagram

[asy] /* Made by MRENTHUSIASM */ size(250); real xMin = -15; real xMax = 15; real yMin = -17; real yMax = 17; draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label("$x$",(xMax,0),(2,0)); label("$y$",(0,yMax),(0,2)); real f(real x) { return x^2+2*x-15; } draw(graph(f,-6.75,4.75),red); pair A, B, C; A = (-5,0); B = (0,-15); C = (3,0); draw(A--B--C); dot("$A$",A,1.5NW,linewidth(4.5)); dot("$B$",B,1.5SE,linewidth(4.5)); dot("$C$",C,1.5NE,linewidth(4.5)); label("$y=x^2+2x-15$",(12,9),red); [/asy] ~MRENTHUSIASM

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

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

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 \[\cos(\angle ABC) = \frac{AB^2 + BC^2 - AC^2}{2 AB BC} = \frac{250 + 234 - 64}{30 \sqrt{260}} = \frac{7}{\sqrt{65}}.\] 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

Alternatively, we could observe that $\sin(\angle ABC)=\sqrt{1-\cos^2(\angle ABC)}=\sqrt{1-\dfrac7{\sqrt{65}}}=\sqrt{{16}{65}}=\dfrac4{\sqrt{65}}$, so $\tan(\angle ABC)=\dfrac{\sin(\angle ABC)}{\cos(\angle ABC)}=\boxed{\textbf{(E) }\dfrac47}$.

~Technodoggo

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 formula $[ABC]=\frac{1}{2}ab\sin{C}.$

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

Solution 6 (Complex Numbers)

[asy] /* Made by MRENTHUSIASM */ size(300); real xMin = -15; real xMax = 15; real yMin = -17; real yMax = 17; draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label("$x$",(xMax,0),(2,0)); label("$y$",(0,yMax),(0,2)); real f(real x) { return x^2+2*x-15; } draw(graph(f,-6.75,4.75),red); pair A, B, C, O; A = (-5,0); B = (0,-15); C = (3,0); O = origin; markscalefactor=0.1; draw(rightanglemark(B,O,C)); draw(A--B--C); dot("$A$",A,1.5SW,linewidth(4.5)); dot("$B$",B,1.5SE,linewidth(4.5)); dot("$C$",C,1.5SE,linewidth(4.5)); dot("$O$",O,1.5SW,linewidth(4.5)); label("$y=x^2+2x-15$",(12,9),red); label("$5$",(-2.5,0),1.5N); label("$3$",(1.5,0),1.5N); label("$15$",(0,-6),W); label("$\theta$",(0,-15),9*dir(100)); label("$\phi$",(0,-15),9*dir(84)); [/asy] From $x^2 + 2x - 15 = (x-3)(x+5)$, we may assume, without loss of generality, that $x$-intercepts of the given parabola are $A(-5,0)$ and $C(3,0)$. And, point $B$ has coordinates $(0,-15)$. Consider complex numbers $z = 3 + i$ and $w = 5 + i$ whose arguments are $\theta \coloneqq \angle OBA$ and $\phi \coloneqq \angle OBC$, respectively. Notice that $\angle ABC = \theta + \phi$ is the argument of the product $zw$ which is \[zw = (3+i)(5+i) = 14 + 8i.\] Hence \[\tan \angle ABC = \frac{\operatorname{Im}(zw)}{\operatorname{Re}(zw)} = \frac{8}{14} = \boxed{\textbf{(E)}\ \frac{4}{7}}.\]

~VensL.

Video Solution by mop 2024

https://youtu.be/ezGvZgBLe8k&t=458s

~r00tsOfUnity

Video Solution (Under 2 min!)

https://youtu.be/InJgY_JYBkE

~Education, the Study of Everything

Video Solution(1-16)

https://youtu.be/SCwQ9jUfr0g

~~Hayabusa1

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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