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The problem states:

- \( T \) is in the interior of \( \angle PQR \).
- \( \overrightarrow{QT} \) bisects \( \angle PQR \), meaning that \( m\angle PQT = m\angle TQR \).
- We are given:
  \[
  m\angle RQT = (6x^2)^\circ
  \]
  and
  \[
  m\angle PQT = (11x + 10)^\circ.
  \]

Since \( \overrightarrow{QT} \) bisects \( \angle PQR \), we know that:
\[
m\angle PQT = m\angle TQR.
\]
Thus, we can set the two angle measures equal:
\[
11x + 10 = 6x^2.
\]

### Step 1: Solve for \( x \)
Rearrange the equation:
\[
6x^2 - 11x - 10 = 0.
\]
We can solve this quadratic equation using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},
\]
where \( a = 6 \), \( b = -11 \), and \( c = -10 \).

Let's calculate the values of \( x \).The solutions to the quadratic equation \( 6x^2 - 11x - 10 = 0 \) are:

\[
x_1 = 2.5 \quad \text{and} \quad x_2 = -\frac{2}{3}.
\]

Since angles cannot be negative, we discard \( x_2 = -\frac{2}{3} \) and take \( x_1 = 2.5 \).

### Step 2: Find \( m\angle PQR \)

Since \( \overrightarrow{QT} \) bisects \( \angle PQR \), we know that:
\[
m\angle PQR = m\angle PQT + m\angle RQT.
\]

Substitute \( x = 2.5 \) into the expressions for \( m\angle PQT \) and \( m\angle RQT \):

\[
m\angle PQT = 11x + 10 = 11(2.5) + 10 = 27.5 + 10 = 37.5^\circ,
\]
\[
m\angle RQT = 6x^2 = 6(2.5)^2 = 6(6.25) = 37.5^\circ.
\]

Thus:
\[
m\angle PQR = 37.5^\circ + 37.5^\circ = 75^\circ.
\]

So, \( m\angle PQR = 75^\circ \).

Would you like further details or explanations?

Here are some related questions:
1. What are the steps to solve a quadratic equation using factoring?
2. How is the quadratic formula derived?
3. What is the importance of angle bisectors in geometry?
4. Can a quadratic equation have complex solutions in geometry problems?
5. What are some other methods to solve quadratic equations?

**Tip:** Always check the discriminant when solving quadratic equations to ensure whether real or complex solutions are expected.

T is in the interior of ∠PQR. Find m∠PQR if QT bisects ∠PQR, m∠RQT = (6x²)°, and m∠PQT = (11x + 10)°.

Learn how to find m∠PQR given the quadratic equations for m∠PQT = (11x + 10)° and m∠RQT = (6x²)°, with QT bisecting the angle. This problem explores quadratic equations and the Angle Bisector Theorem, suitable for Grades 9-11.

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