Find the measure of the angle from the data in the triangle

Question

I have to find $\angle AMN$ of the triangle $\triangle ABM$ below.

The triangle $\triangle ABM$ is isosceles. I know that $\angle AMN+\angle BMN=180^{\circ}-2\cdot50^{\circ}=80^{\circ}.$ From the picture, we get $\angle MAN=40^\circ$ and $\angle ABN=30^{\circ}.$ Also, $\angle ANB=180^{\circ}-10^{\circ}-30^{\circ}=140^{\circ}$ in $\triangle ABN.$ But I don't know how to find $\angle AMN.$ What can I use to find it?

Answer 1

Here is another way to solve the exercise.

Since the triangles $\,AEK\,$ and $\,AEH\,$ are congruent by ASA Criterion, it results that $\,EK\cong AH\,.$

Moreover the triangles $\,ABM\,,\,BME\,$ and $\,ANE\,$ are isosceles and the triangle $\,BEK\,$ is half an equilateral triangle, consequently,

$AM\cong BM\cong BE\cong 2EK\cong 2AH\cong AN\;\;,$

hence,

the triangle $\,AMN\,$ is isosceles too and

$\angle AMN=\dfrac{180\unicode{176}-\angle MAN}2=70\unicode{176}\,.$

Answer 2

EDIT. I posted this answer too quickly, but I'll leave it up as in incomplete answer. As noted in the comments, this system is underdetermined, so there is a $1$-parameter infinite family of solutions. (The matrix associated to the system has rank $3$.) Hence, we need one more piece of information to determine the angles.

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Let's give shorter names to four of the angles: \begin{align} w &= \angle AMN \\ x &= \angle BMN \\ y &= \angle ANM \\ z &= \angle BNM \end{align}

Then using the fact that the sum of all three angles in a planar triangle is $180^\circ$ and the sum of all angles around a point is $360^\circ$, we can write down four linear equations. Try it yourself before revealing the spoiler.

$$\left\{ \begin{aligned} w + x &= 80 \\ y + z &= 220 \\ w + y &= 140 \\ x + z &= 160 \end{aligned} \right.$$

Hopefully, you can take it from here.

Answer 3

Here are two solutions.

Synthetic geometry

Let $P$ be the circumcenter of $\triangle ABN$. Since $$ \angle APN = 2 \angle ABN = 60^\circ, $$ we have $AN = AP$. Since $$ \angle APB = 2 (180^\circ - \angle ANB) = 80^\circ, $$ the quadrilateral $AMBP$ is a rhombus, so $AP = AM$. Therefore, $AN = AM$, so $$ \angle AMN = \frac{180^\circ - \angle MAN}{2} = 70^\circ. $$

Trigonometry

Assume that $AM = BM = 1$, so $AB = 2 \sin 40^\circ$. Applying the law of sines in $\triangle ABN$ shows that $$ AN = 2 \sin 40^\circ \cdot \frac{\sin 30^\circ}{\sin 140^\circ} = 1, $$ so $$ \angle AMN = \frac{180^\circ - \angle MAN}{2} = 70^\circ. $$