Difference between revisions of "2023 AMC 12A Problems/Problem 15"

(Solution 1)
Line 1: Line 1:
 
==Question==
 
==Question==
  
[uh someone insert diagram]
+
Usain is walking for exercise by zigzagging across a <math>100</math>-meter by <math>30</math>-meter rectangular field, beginning at point <math>A</math> and ending on the segment <math>\overline{BC}</math>. He wants to increase the distance walked by zigzagging as shown in the figure below <math>(APQRS)</math>. What angle <math>\theta</math><math>\angle PAB=\angle QPC=\angle RQB=\cdots</math> will produce in a length that is <math>120</math> 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.)
 +
 
 +
<math>\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}</math>
  
 
==Solution 1==
 
==Solution 1==
Line 10: Line 12:
  
 
<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==
 +
https://youtu.be/NhUI-BNCIUE
 +
  
 
==See also==
 
==See also==

Revision as of 22:49, 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 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 (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png