To solve for the angles in the isosceles triangle $\triangle ABC$, where base $CA$ is given, and the angles $m\angle B$ and $m\angle C$ are expressed as $2x + 36$ and $5x - 6$, respectively, follow these steps:

Step 1: Recall the triangle angle sum property.

The sum of all angles in a triangle equals $180^\circ$: $m\angle A + m\angle B + m\angle C = 180^\circ.$

Step 2: Use the isosceles property.

Since $\triangle ABC$ is isosceles with base $CA$, the two base angles $m\angle B$ and $m\angle C$ are congruent: $m\angle B = m\angle C.$ From the problem: $m\angle B = 2x + 36 \quad \text{and} \quad m\angle C = 5x - 6.$ Thus: $2x + 36 = 5x - 6.$

Step 3: Solve for $x$.

Simplify the equation: $2x + 36 = 5x - 6.$ Subtract $2x$ from both sides: $36 = 3x - 6.$ Add 6 to both sides: $42 = 3x.$ Divide by 3: $x = 14.$

Step 4: Find $m\angle B$ and $m\angle C$.

Substitute $x = 14$ into the expressions for $m\angle B$ and $m\angle C$: $m\angle B = 2x + 36 = 2(14) + 36 = 28 + 36 = 64^\circ,$ $m\angle C = 5x - 6 = 5(14) - 6 = 70 - 6 = 64^\circ.$

Step 5: Find $m\angle A$.

Using the angle sum property: $m\angle A + m\angle B + m\angle C = 180^\circ.$ Substitute $m\angle B = 64^\circ$ and $m\angle C = 64^\circ$: $m\angle A + 64 + 64 = 180.$ Simplify: $m\angle A = 180 - 128 = 52^\circ.$

Final Answer:

The degree measures of the angles are: $m\angle A = 52^\circ, \quad m\angle B = 64^\circ, \quad m\angle C = 64^\circ.$

Do you want further clarification or additional steps? Here are 5 related questions:

1. How would the solution change if the triangle were equilateral?
2. What if $m\angle B$ and $m\angle C$ were not congruent? How would we solve for $x$?
3. Can you generalize the angle sum property for a quadrilateral?
4. What happens to the angle measures if we increase $x$ further?
5. How do you verify that the triangle satisfies the isosceles property?

Tip: Always verify that the calculated angles add up to $180^\circ$ when solving for angles in any triangle.