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2023 AMC 12A Problems/Problem 15: Difference between revisions

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==Solution 1==
==Solution 1==


By "unfolding" line APQRS into a straight line, we get a right triangle ABS.
By "unfolding" line <math>APQRS</math> into a straight line, we get a right angled triangle <math>ABS</math>.


<math>cos(\theta)=\frac{120}{100}</math>
<math>cos(\theta)=\frac{120}{100}</math>


<math>\theta=\boxed{\textbf{(A) } cos^-1(\frac{5}{6})}</math>
<math>\theta=\boxed{\textbf{(A) } cos^-1(\frac{5}{6})}</math>


==Video Solution 1 by OmegaLearn==
==Video Solution 1 by OmegaLearn==

Revision as of 22:51, 9 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.)

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

Solution 1

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

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

$\theta=\boxed{\textbf{(A) } cos^-1(\frac{5}{6})}$

Video Solution 1 by OmegaLearn

https://youtu.be/NhUI-BNCIUE


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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