To solve for $x$ and the measure of angle $C$ in this isosceles triangle, let's go through the problem step by step.

Step 1: Understanding the Triangle's Properties

1. This is an isosceles triangle, which means two sides are equal, and hence the angles opposite these sides are also equal.
2. Given:
   • $\angle A = 80^\circ$,
   • Side $AB = 4y - 2$,
   • Side $AC = 2y + 2$,
   • Angle $\angle C = (2x)^\circ$.

Since $AB = AC$, we can use this information to find $y$ first.

Step 2: Set Up the Equation Using Equal Sides

Since $AB = AC$, $4y - 2 = 2y + 2.$

Solve for $y$:

$4y - 2y = 2 + 2,$ $2y = 4,$ $y = 2.$

Step 3: Using $y = 2$, Confirm Equal Sides

Substitute $y = 2$ back into the expressions for $AB$ and $AC$: $AB = 4(2) - 2 = 8 - 2 = 6,$ $AC = 2(2) + 2 = 4 + 2 = 6.$ Thus, $AB = AC$, confirming that the triangle is indeed isosceles.

Step 4: Use Angle Sum in a Triangle to Find $x$

The sum of angles in any triangle is $180^\circ$. Therefore, $\angle A + \angle B + \angle C = 180^\circ.$ Since $\angle B = \angle C$ (due to the isosceles property), let $\angle B = \angle C = (2x)^\circ$.

So, $80^\circ + 2x + 2x = 180^\circ,$ $80^\circ + 4x = 180^\circ.$

Solve for $x$:

$4x = 180^\circ - 80^\circ,$ $4x = 100^\circ,$ $x = 25^\circ.$

Step 5: Find the Measure of Angle $C$

Since $\angle C = (2x)^\circ$, $\angle C = 2 \cdot 25^\circ = 50^\circ.$

Final Answer

• $x = 25^\circ$.
• The measure of angle $C = 50^\circ$.

Would you like further clarification on any of the steps?

Here are 5 related questions to deepen your understanding:

1. What are the steps to solve for $y$ if $AB$ and $AC$ had different expressions?
2. How does the isosceles property help in solving this type of triangle problem?
3. What would change if angle $A$ were different from $80^\circ$?
4. How do we verify our answer by recalculating the angles?
5. How would the solution differ if $\angle B$ and $\angle C$ were not equal?

Tip: Always double-check that all triangle angles add up to $180^\circ$ to confirm the correctness of your solution.