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Difference between revisions of "2023 AMC 12A Problems/Problem 15"
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Now we have: <cmath>\cos\theta = \frac{10\sin\theta - 9\cos\theta}{12\sin\theta - 9},</cmath> meaning that: <cmath>12\sin\theta\cos\theta - 9\cos\theta = 10\sin\theta - 9\cos\theta \implies \cos\theta = \frac{10}{12} = \frac56.</cmath>
 
Now we have: <cmath>\cos\theta = \frac{10\sin\theta - 9\cos\theta}{12\sin\theta - 9},</cmath> meaning that: <cmath>12\sin\theta\cos\theta - 9\cos\theta = 10\sin\theta - 9\cos\theta \implies \cos\theta = \frac{10}{12} = \frac56.</cmath>
 
This means that <math>\theta = \arccos\left(\frac56\right)</math>, giving us <math>\boxed{\textbf{A}}</math>
 
This means that <math>\theta = \arccos\left(\frac56\right)</math>, giving us <math>\boxed{\textbf{A}}</math>
 +
 +
~ap246
  
 
==See also==
 
==See also==

Revision as of 10:32, 10 November 2023

Question

Usain is walking for exercise by zigzagging across a $100$-meter by $30$-meter rectangular field, beginning at point $A$ and ending on the segment $\overline{BC}$. He wants to increase the distance walked by zigzagging as shown in the figure below $(APQRS)$. What angle $\theta$$\angle PAB=\angle QPC=\angle RQB=\cdots$ will produce in a length that is $120$ meters? (This figure is not drawn to scale. Do not assume that he zigzag path has exactly four segments as shown; there could be more or fewer.)

[someone add diagram]

$\textbf{(A)}~\arccos\frac{5}{6}\qquad\textbf{(B)}~\arccos\frac{4}{5}\qquad\textbf{(C)}~\arccos\frac{3}{10}\qquad\textbf{(D)}~\arcsin\frac{4}{5}\qquad\textbf{(E)}~\arcsin\frac{5}{6}$

Solution 1

By "unfolding" $APQRS$ into a straight line, we get a right angled triangle $ABS$.

$cos(\theta)=\frac{100}{120}$

$\theta=\boxed{\textbf{(A) } \arccos\left(\frac{5}{6}\right)}$

~lptoggled

Video Solution 1 by OmegaLearn

https://youtu.be/NhUI-BNCIUE


Solution 2(Trig Bash)

We can let $x$ be the length of one of the full segements of the zigzag. We can then notice that $\sin\theta = \frac{30}{x}$. By Pythagorean Theorem, we see that $BP = \sqrt{x^2 - 900}$. This implies that: \[RC = 100 - 3\sqrt{x^2 - 900}.\] We also realize that $RS = 120 - 3x$, so this means that: \[\cos\theta = \frac{100 - 3\sqrt{x^2 - 900}}{120 - 3x}.\] We can then substitute $x = \frac{30}{\sin\theta}$, so this gives: \begin{align*} \cos\theta &= \frac{100 - 3\sqrt{x^2 - 900}}{120 - 3x}\\ &= \frac{100 - 3\sqrt{\frac{900}{\sin^2\theta} - 900}}{120 - \frac{90}{\sin\theta}}\\ &= \frac{100 - 90\sqrt{\csc^2\theta - 1}}{120 - \frac{90}{\sin\theta}}\\ &= \frac{100 - \frac{90}{\tan\theta}}{120 - 90\sin\theta}\\ &= \frac{100\sin\theta - 90\cos\theta}{120\sin\theta - 90}\\ &= \frac{10\sin\theta - 9\cos\theta}{12\sin\theta - 9}\\ \end{align*}

Now we have: \[\cos\theta = \frac{10\sin\theta - 9\cos\theta}{12\sin\theta - 9},\] meaning that: \[12\sin\theta\cos\theta - 9\cos\theta = 10\sin\theta - 9\cos\theta \implies \cos\theta = \frac{10}{12} = \frac56.\] This means that $\theta = \arccos\left(\frac56\right)$, giving us $\boxed{\textbf{A}}$

~ap246

See also

2023 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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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