To solve the problem, we need to find the measure of angle $m\angle SQR$. We are given the following information about the angles in triangle $\Delta QRS$ and the extended line segment:

• $m\angle RST = (6x + 7)^\circ$
• $m\angle QRS = (x - 3)^\circ$
• $m\angle SQR = (3x + 20)^\circ$

In any triangle, the sum of the interior angles is always $180^\circ$. Therefore, in triangle $\Delta QRS$, we have:

$m\angle QRS + m\angle SQR + m\angle QSR = 180^\circ$

But we are not given $m\angle QSR$ directly. However, $m\angle RST$ is the exterior angle at point $S$ which is equal to the sum of the two opposite interior angles, $m\angle SQR$ and $m\angle QRS$:

$m\angle RST = m\angle SQR + m\angle QRS$

Substitute the given expressions:

$6x + 7 = (3x + 20) + (x - 3)$

Now, let's solve for $x$:

$6x + 7 = 3x + x + 20 - 3$ $6x + 7 = 4x + 17$ $6x - 4x = 17 - 7$ $2x = 10$ $x = 5$

Now that we have $x = 5$, substitute this value back into the expression for $m\angle SQR$:

$m\angle SQR = 3x + 20 = 3(5) + 20 = 15 + 20 = 35^\circ$

So, the measure of $m\angle SQR$ is $35^\circ$.

Would you like more details or have any questions?

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Here are five related questions:

1. What is the measure of angle $m\angle QRS$?
2. What is the measure of angle $m\angle RST$?
3. What is the sum of the angles in triangle $\Delta QRS$?
4. How does the exterior angle theorem help in solving this problem?
5. Can you verify the solution by finding $m\angle QSR$?

Tip: Always check the relationships between interior and exterior angles when working with triangles extended to form exterior angles.