Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 3.7 AMC 10/12 Find the answer

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 $PA?$

Pick one

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 $N$ is the midpoint of $U$ and $G$ , or $\left(\frac{10+\cos(36)-\cos(56)}{2}, \frac{\sin(36)+\sin(56)}{2}\right)$
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 $\frac{\sin(36)+\sin(56)}{\cos(36)-\cos(56)}$
Now, using sum to product identities, we have this equal to $\frac{2\sin(46)\cos(10)}{-2\sin(46)\sin({-10})}=\frac{\sin(80)}{\cos(80)}=\tan(80)$
so the answer is $\boxed{\textbf{(E) } 80}.$
~lifeisgood03
Note: Though this solution is excellent, setting $M = (0,0)$ makes life a tad bit easier
~MathleteMA

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.