# In $\Delta ABC$, $\angle C = 36^{\circ}$ and $\angle B = 96^{\circ}$. What is $\angle MNC$?

In $$\Delta ABC$$, $$\angle C = 36^{\circ}$$ and $$\angle B = 96^{\circ}$$. $$N$$ lies on $$AC$$ such that $$AN = NE$$, where $$E$$ is defined such that $$CE = AB$$. $$M$$ is the midpoint of $$BC$$. What is $$\angle MNC$$?

I got this question from a friend on Discord who also was stuck. So I drew the diagram on GeoGebra and the answer turns out to be $$24^{\circ}$$ (https://www.geogebra.org/geometry/cmdxcdds). The condition $$CE = AB$$ really stood out to me, so I defined point $$D$$ such that $$AB = BD$$, and it turns out that $$BD \perp NM$$. I have no idea how to prove this however, and I haven't had any success with using other theorems such as the angle in the circle being twice the angle on the circumference.

I would preferably like a solution involving only Euclidean geometry, but a solution with trigonometry is fine.

• Hint: note that vector $\overline{MN}$ is the halfsum of vectors $\overline{BA}$ and $\overline{CE}$ which have the same length. Can you conclude now? Nov 26, 2023 at 18:13

Let $$R$$ be the midpoint of $$AC$$. Then $$MR \cong \frac12 AB.$$ We also have $$ER \cong AB -\frac12AC\tag{1}\label{1}$$ and $$AN \cong NE \cong \frac12(AC-AB).\tag{2}\label{2}$$Adding \eqref{1} and \eqref{2} we conclude that $$MNR$$ is isosceles. Since $$\measuredangle MRC = 48^\circ$$, form Exterior Angle Theorem we reach our result $$\boxed{\measuredangle MNC = 24^\circ}.$$

Alternative solution:
I will use sine rule too, but for both triangles $$\triangle ABC$$ and $$\triangle MNC$$, and well known relation: $$\sin \theta = \sin (180^{\circ} - \theta)$$. But first let find $$CN$$!

We know $$CE = AB$$, so $$AE = AC - AB \implies NE = \dfrac{AC-AB}{2}$$. Now, we have: $$CN=CE+NE=AB+\frac{AC-AB}{2}=\frac{AB+AC}{2}.$$ If we use sine rule for triangle $$\triangle ABC$$, we have: $$\frac{AB}{\sin 36^{\circ}}=\frac{AC}{\sin 96^{\circ}}=\frac{BC}{\sin 48^{\circ}}. \qquad (1)$$ If we use sine rule for triangle $$\triangle MNC$$, we have: $$\frac{\dfrac{AC+AB}{2}}{\sin(180^{\circ}-(36^{\circ}+\theta))}=\frac{\dfrac{BC}{2}}{\sin \theta}. \qquad (2)$$ From $$(1)$$ we have: $$AC=\frac{BC\sin 96^{\circ}}{\sin 48^{\circ}}, \quad AB=\frac{BC\sin 36^{\circ}}{\sin 48^{\circ}},$$ hence we can rewrite equation $$(2)$$ as: $$\frac{\sin96^{\circ}+\sin 36^{\circ}}{\sin 48^{\circ} \sin(36^{\circ}+\theta)}=\frac{1}{\sin\theta} \iff (\sin96^{\circ}+\sin 36^{\circ})\sin\theta=\sin 48^{\circ} \sin(36^{\circ}+\theta),$$ or $$(\sin96^{\circ}+\sin 36^{\circ})\sin\theta=\sin48^{\circ}\sin36^{\circ}\cos\theta+\sin48^{\circ}\cos36^{\circ}\sin\theta,$$ or $$\tan\theta = \frac{\sin48^{\circ}\sin36^{\circ}}{\sin96^{\circ}+\sin 36^{\circ}-\sin48^{\circ}\cos36^{\circ}} \implies \boxed{\theta = 24^{\circ}}.$$

Here is a brute-force answer using trigonometry. By the sine rule:

$$\frac{AC}{\sin 96^{\circ}} = \frac{BC}{\sin 48^{\circ}} = \frac{AB}{\sin 36^{\circ}}$$

Without loss of generality, let $$AC = 1$$. Then $$BC = \frac{\sin 48^{\circ}}{\sin 96^{\circ}}, AB = \frac{\sin 36^{\circ}}{\sin 96^{\circ}}$$. Now let us find the coordinates of $$M$$ using coordinate geometry, where $$A$$ is located at the origin. $$B = (AB \cos 48^{\circ}, AB \sin 48^{\circ}) = \left(\frac{\sin 36^{\circ} \cos 48^{\circ}}{\sin 96^{\circ}}, \frac{\sin 36^{\circ} \sin 48^{\circ}}{\sin 96^{\circ}} \right)$$. Hence $$M = \left(\frac{1}{2} + \frac{\sin 36^{\circ} \cos 48^{\circ}}{2 \sin 96^{\circ}}, \frac{\sin 36^{\circ} \sin 48^{\circ}}{2\sin 96^{\circ}} \right)$$. Also, $$E = (1 - \frac{\sin 36^{\circ}}{\sin 96^{\circ}}, 0)$$ so $$N = \left(\frac{1}{2} - \frac{\sin 36^{\circ}}{2 \sin 96^{\circ}}, 0 \right)$$.

Thus if $$x = \angle MNC$$:

$$\tan x = \frac{\sin 36^{\circ} \sin 48^{\circ}}{\sin 36^{\circ} \cos 48^{\circ}+\sin 36^{\circ}}$$ $$= \frac{\sin 48^{\circ}}{\cos 48^{\circ} + 1}$$

where by the tangent half-angle formula, we recognise this as $$\tan 24^{\circ}$$. Hence $$\angle MNC = 24^{\circ}$$.