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The problem involves two isosceles triangles, \( \triangle PQR \) and \( \triangle OQR \), and we are given the relation that \( \angle a = 2 \times \angle b \), and we need to find \( \angle c \). Let's proceed step by step:

### Step 1: Using properties of isosceles triangles
- In \( \triangle PQR \), since it is isosceles with \( PQ = PR \), the base angles are equal: 
  \[
  \angle PQR = \angle PRQ = b
  \]
  The sum of the angles in any triangle is \( 180^\circ \), so:
  \[
  \angle P + 2b = 180^\circ 
  \]
  Which gives us:
  \[
  \angle P = 180^\circ - 2b
  \]

- In \( \triangle OQR \), which is also isosceles with \( OQ = OR \), the base angles are:
  \[
  \angle OQR = \angle ORQ = a
  \]
  And similarly, the sum of angles in \( \triangle OQR \) is \( 180^\circ \), so:
  \[
  \angle O + 2a = 180^\circ
  \]
  Which gives:
  \[
  \angle O = 180^\circ - 2a
  \]

### Step 2: Using the given relation \( \angle a = 2 \times \angle b \)
We are told that \( a = 2b \), so substitute \( a = 2b \) into the equation for \( \angle O \):
\[
\angle O = 180^\circ - 2(2b) = 180^\circ - 4b
\]

### Step 3: Finding \( \angle c \)
We know from the diagram that \( \angle c \) is the external angle at vertex \( R \), formed by the sum of \( \angle OQR \) and \( \angle O \). Thus:
\[
\angle c = \angle OQR + \angle O = a + (180^\circ - 4b)
\]
Substituting \( a = 2b \):
\[
\angle c = 2b + (180^\circ - 4b) = 180^\circ - 2b
\]

### Step 4: Using the given angle \( \angle Q = 20^\circ \)
Finally, it is given that \( \angle Q = 20^\circ \). Since \( \angle c = 20^\circ \), we now know that \( \angle c = 20^\circ \).

Thus, the value of \( \angle c \) is **20 degrees**.

Would you like further clarification or more details? Here are some related questions to expand on this topic:

1. What is the general formula for finding angles in isosceles triangles?
2. How does the relationship between angles affect the triangle's shape?
3. How can external angles help solve problems involving multiple triangles?
4. What happens when the base angles are not equal in a triangle?
5. How can we apply the sum of angles to non-triangular polygons?

**Tip**: Always remember that the sum of angles in any triangle is 180°, a fundamental rule in geometry.

In the following figure, ∆PQR and ∆OQR are isosceles triangles and ∠a = 2∠b. Find ∠c.

Learn how to solve a geometric problem involving two isosceles triangles and the relationship between their angles. This example shows how to apply the isosceles triangle theorem and the external angle theorem to find the missing angle ∠c, given that ∠a = 2∠b.

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