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Difference between revisions of "2018 AMC 12A Problems/Problem 23"

(Created page with "=== Problem === In <math>\triangle PAT,</math> <math>\angle P=36^{\circ},</math> <math>\angle A=56^{\circ},</math> and <math>PA=10.</math> Points <math>U</math> and <math>G<...")
 
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=== Problem ===
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== Problem ==
  
  
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\textbf{(E)} 80 </math>
 
\textbf{(E)} 80 </math>
  
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=== Solution ===
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== Solution ==
  
 
Let <math>P</math> be the origin, and <math>PA</math> lie on the x axis.
 
Let <math>P</math> be the origin, and <math>PA</math> lie on the x axis.

Revision as of 14:39, 8 February 2018

Problem

In $\triangle PAT,$ $\angle P=36^{\circ},$ $\angle A=56^{\circ},$ and $PA=10.$ Points $U$ and $G$ lie on sides $\overline{TP}$ and $\overline{TA},$ respectively, so that $PU=AG=1.$ Let $M$ and $N$ be the midpoints of segments $\overline{PA}$ and $\overline{UG},$ respectively. What is the degree measure of the acute angle formed by lines $MN$ and $PA?$

$\textbf{(A)} 76 \qquad \textbf{(B)} 77 \qquad \textbf{(C)} 78 \qquad \textbf{(D)} 79 \qquad \textbf{(E)} 80$

Solution

Let $P$ be the origin, and $PA$ lie on the x axis.

We can find $U=\left(\cos(36), \sin(36)\right)$ and $G=\left(10-\cos(56), \sin(56)\right)$

Then, we have $M=(5, 0)$ and $N=\left(\frac{10+\cos(36)-\cos(56)}{2}, \frac{\sin(36)+\sin(56)}{2}\right)$

Notice that the tangent of our desired points is the the absolute difference between the y coordinates of the two points divided by the absolute difference between the x coordinates of the two points.

This evaluates to \[\frac{\cos(36)-\cos(56)}{\sin(36)+\sin(56)}\] Now, using sum to product identities, we have this equal to \[\frac{2\cos(10)\cos(46)}{2\sin(10)\cos(46)}=\tan(80)\] so the answer is $\boxed{\textbf{(E)}.}$ (lifeisgood03)

See Also

2018 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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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