Since quadrilateral $QRST$ is inscribed in a circle, it is a cyclic quadrilateral. This implies that opposite angles are supplementary, meaning their measures add up to $180^\circ$.

Given:

• $m\angle Q = 68^\circ$
• $m\angle R = (3x + 40)^\circ$
• $m\angle T = (5x - 52)^\circ$

Since $\angle Q$ and $\angle S$ are opposite angles, they should add up to $180^\circ$: $m\angle Q + m\angle S = 180^\circ$ Thus, $68^\circ + m\angle S = 180^\circ$ $m\angle S = 180^\circ - 68^\circ = 112^\circ$

Now, $\angle R$ and $\angle T$ are also opposite angles and should add up to $180^\circ$: $(3x + 40) + (5x - 52) = 180$ Simplifying this equation: $8x - 12 = 180$ $8x = 192$ $x = 24$

Now, substitute $x = 24$ into the expressions for $m\angle R$ and $m\angle T$: $m\angle R = 3(24) + 40 = 72 + 40 = 112^\circ$ $m\angle T = 5(24) - 52 = 120 - 52 = 68^\circ$

Therefore, the measures are: $m\angle R = 112^\circ, \; m\angle S = 112^\circ, \; m\angle T = 68^\circ$

Would you like to see the step-by-step process in more detail?

1. How does the cyclic property of quadrilaterals affect angle relationships?
2. How can we verify if a quadrilateral is cyclic?
3. What are some real-world applications of cyclic quadrilaterals?
4. Could this approach be applied to non-cyclic quadrilaterals?
5. Why is it important that $QRST$ is inscribed in circle $P$?

Tip: When working with cyclic quadrilaterals, remember that opposite angles will always add up to $180^\circ$.