Consider the triangle $PAT$, with angle $P = 36$ degres, angle $A = 56$ degrees and $PA=10$. The points $U$ and $G$ lie on sides TP and TA respectively, such that PU = AG = 1. Let M and N be the midpoints of segments PA and UG. What is the degree measure of the acute angle formed by the lines MN and PA?
It would be very helpful if anyone had a solution using complex numbers to this problem.
On
A solution using complex numbers was requested, here is a possibility to write it down. First of all, let us find some good places (affixes) for the given points. I will use a small letter for the affix of the point with the corresponding capital letter. (So $a\in\Bbb C$ is the affix of the point $A$ in the plane, et caetera.)
Since $PA=10$, let us arrange $p=-5$, and $a=5$. The mid point $M$ of $PA$ corresponds to $\frac 12(p+a)=0$.
The complex numbers $u,g,n$ are then $$ \begin{aligned} u &= -5 + 1\cdot (\cos 36^\circ +i\sin 36^\circ)\ ,\\ g &= +5 + 1\cdot (\cos 124^\circ +i\sin 124^\circ)\ ,\\ n &= \frac 12(u+g) % &=\frac 12 % \Big( \ % (\cos 36^\circ +i\sin 36^\circ) % + % (\cos 124^\circ +i\sin 124^\circ)\ \Big) \\ &= \frac 12(\cos 36^\circ+\cos 124^\circ) + \frac i2(\sin 36^\circ+\sin 124^\circ) \\ &= \cos\frac{36^\circ+124^\circ}2 \cos\frac{36^\circ-124^\circ}2 + i \sin\frac{36^\circ+124^\circ}2 \cos\frac{36^\circ-124^\circ}2 \\ &= \cos 44^\circ\Big(\ \cos 80^\circ+i\sin 80^\circ \ \Big)\ . \end{aligned} $$ So the needed angle is $80^\circ$, extracted from the above trigonometric form of $n$.
The length $10$ of the base is a red herring; we find that it cancels out, no matter what its value is.
We are effectively asked to find the midpoint of $\text{cis }36$ and $\text{cis }(180-56) = \text{cis }124$, and take this midpoint's argument. This is, of course, halfway between $36$ and $124$, which is $\frac{36+124}{2} = 80$.