To solve this problem, let's analyze the given geometric figure and the associated relationships:

Given:

• PQRS is a cyclic quadrilateral (which means opposite angles are supplementary).
• $QS = QR$ (isosceles triangle QSR).
• $VST$ is a tangent to the circle at point $S$.

Angles:

• Angle $\angle PQS = 7x$.
• Angle $\angle PRS = 2x$.

Steps:

1. Isosceles Triangle QSR: Since $QS = QR$, triangle $QSR$ is isosceles. This means that the base angles $\angle QSR$ and $\angle QRS$ are equal.

2. Cyclic Quadrilateral Property: In a cyclic quadrilateral, opposite angles sum to 180°. Therefore, $\angle PQS + \angle PRS = 180^\circ.$ Substituting the values: $7x + 2x = 180^\circ.$ Simplifying: $9x = 180^\circ,$ $x = 20^\circ.$

Thus, the value of $x$ is 20°.

Would you like more details on this solution?

Relative Questions:

1. How does the cyclic quadrilateral property help solve such problems?
2. What is the significance of the isosceles triangle in this context?
3. How does the tangent line influence angle relations in the circle?
4. What are the key properties of cyclic quadrilaterals?
5. Can you explain the supplementary angle concept further?

Tip:

Always check for isosceles triangles or cyclic quadrilateral properties in circle-related geometry problems—these often provide key angle relationships!