Calculate the measure in degrees of the acute angle formed by $ NMA $?

Question

Consider the $PAT$ triangle such that the angle $ \angle P = 36 $ degrees, $ \angle A = 56 $ degrees and $ PA = 10 $. Knowing that the points $ U $ and $ G $ belong, respectively, to the sides $ TP $ and $ TA $, so that $ PU = AG = 1 $. If $ M $ and $ N $ are the midpoints of the $ PA $ and $ UG $ segments, respectively. What is the measure in degree of the acute angle formed by $ NMA $?

Can anyone give me any tips?

Answer 1

Hint :

Use complex numbers. Let affixes of $A,P$ be $z_A=-5$, $z_P=5$. So $z_M=0$.

$z_G$ would be $-5+e^{i\,56^{\circ}}$. Why? Similarly find affixes of $U,N$.

Argument of $z_N$ will help you compute $\angle NMA$. I got it as a nice integer.

Answer 2

Connect $MN$ and $MU$. Rotate $\bigtriangleup MPU$ $\;180°$ about $M$ and name the exterior point $U'$.

Connect $U'G$. Observe that $\bigtriangleup AGU'$ is isosceles and $U'G \; \parallel MN$.

The rest is simple angle chase.

Answer 3

Draw $PG$ and let its midpoint be $D$. Draw $MN$, $MD$ and $ND$.

Observe that, $MD\parallel AG$ and $DN\parallel PU$. Also, $\triangle MDN$ is isosceles because $MD=\frac{1}{2} AG=\frac{1}{2} PU=DN$.

Now, $\angle MDN=\angle MDG+\angle GDN=\angle PGT+\angle GPT=92^{\circ}$

Thus, $\angle NMA=180-\angle NMD-\angle DMP=180-44-56=\boxed{80}$

Answer 4

If you construct $P^*$ and $A^*$ so that $$\,\,\,PP^* \, || \, UN \, , \,\,\,PP^* = UN = \frac{1}{2} UG\,\,\,v\text{ and }\,\,\,PP^* \, || \, UN\, , \,\, PP^* = UN = \frac{1}{2} UG\,\,\,$$ then $UPP^*N$ and $GAA^*N$ are parallelograms and so $$\,\,\,NP^* \, || \, UP \, , \,\,\,NP^* = UP \,\,\, \text{and} \,\,\,NA^* \, || \, GA\, , \,\, NA^* = GA\,\,\,$$ which yields $NP^* = NA^*$ and thus triangle $\Delta \, NP^*A^* $ is isosceles. Furthermore, since $NP^* \, || \, TP$ and $NA^* \, || \, TA$ $$\angle \, P^*NA^* = \angle \, PTA = 88^{\circ}$$ Observe that by construction, $PP^* \, || \, UG\,, \,\,\,AA^* \, || \, UG\,\,$ and $\,\, PP^* = \frac{1}{2}UG = AA^*$, which implies that $PP^*AA^*$ is a parallelogram. By the fact that the diagonals of a parallelogram share a common midpoint, since $M$ is the midpoint of $PA$, it follows that $M$ is also the midpoint of $P^*A^*$. That implies that $NM$ is a median in the isosceles triangle $\Delta \, NP^*A^*$ which means that $MN$ is also the angle bisector of $\angle \, P^*NA^*$ so $$\angle \, MNA^* = \frac{1}{2} \angle \, P^*NA^* = 44^{\circ}$$ Set $L = NA^* \,\cap \, PA$ and since $NA^* \, || \, TA$ $$\angle \, NLM = \angle \, TAP = 56^{\circ}$$ We know two of the three angles od the triangle $\Delta \, NML$ so the thrid angle is
$$\angle \, NML = 180^{\circ} - 56^{\circ} - 44^{\circ} = 80^{\circ}$$