The problem in the image asks to calculate the angle $x$, given the following conditions:

• The triangle is divided into two smaller triangles by the height.
• $\overline{AB} = \overline{AC}$, meaning that triangle $ABC$ is isosceles.
• $\overline{AM} = \overline{MC}$, meaning that $M$ is the midpoint of $\overline{BC}$.
• The angle $\angle AMC = \overline{AM}C$, indicating that angle $AMC$ and $AMB$ are congruent.
• The angle $\angle BMA = 20^\circ$.

Let me proceed with solving the problem.

Step 1: Set up the known angles and relationships.

• Since $\triangle ABC$ is isosceles, we know that $\angle ABC = \angle ACB$.
• The problem states $M$ is the midpoint, so triangles $\triangle AMB$ and $\triangle AMC$ are congruent.
• $\angle BMA = 20^\circ$.

Step 2: Calculate angle $x$.

Since the sum of angles in a triangle is always $180^\circ$, we can calculate $\angle A$ and use symmetry of the isosceles triangle to find angle $x$.

I'll now calculate the angles.The angle $x$, which represents $\angle BAC$, is $90^\circ$.

If you need further details or have any questions, feel free to ask!

Here are 5 related questions for further exploration:

1. What properties of an isosceles triangle were used to solve for the angle $x$?
2. How does the midpoint $M$ divide the triangle $ABC$?
3. What is the significance of the equal angles in triangle $ABC$?
4. Can you prove that the sum of angles in any triangle is always $180^\circ$?
5. How would the solution change if the given angle $\angle BMA$ were different?

Tip: Always check the properties of triangles (e.g., isosceles, equilateral) as they often simplify complex angle calculations.