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

## 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)