Apply the Alternate Segment Theorem.

The Alternate Segment Theorem states that the angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment.

• The tangent is PQ and the chord is AT. The angle between them is $\angle PTA = 60^\circ$. Therefore, the angle in the alternate segment, $\angle ABT$, is equal to $\angle PTA$.

$$\angle ABT = \angle PTA = 60^\circ$$

• The tangent is PQ and the chord is BT. The angle in the alternate segment is $\angle BAT = 44^\circ$. Therefore, the angle between the tangent and the chord, $\angle QTB$, is equal to $\angle BAT$.

$$\angle QTB = \angle BAT = 44^\circ$$

Given that TC bisects $\angle BTQ$, it means that $\angle BTC = \angle CTQ$.
Since $\angle BTQ = \angle QTB = 44^\circ$:

$$\angle BTC = \frac{1}{2} \times \angle BTQ$$ $$\angle BTC = \frac{1}{2} \times 44^\circ$$ $$\angle BTC = 22^\circ$$

The line ABC is a straight line. This means that $\angle ABT$ is an exterior angle to $\triangle BTC$ at vertex B.
The Exterior Angle Theorem states that an exterior angle of a triangle is equal to the sum of the two opposite interior angles.
For $\triangle BTC$, the exterior angle at B is $\angle ABT$, and the opposite interior angles are $\angle BTC$ and $\angle BCT$.

$$\angle ABT = \angle BTC + \angle BCT$$

Substitute the values we found:

$$60^\circ = 22^\circ + \angle BCT$$

Subtract $22^\circ$ from both sides:

$$\angle BCT = 60^\circ - 22^\circ$$ $$\angle BCT = 38^\circ$$

Since $\angle BCT$ is the same as $\angle ACT$, we have:

$$\angle ACT = 38^\circ$$

The final answer is \boxed{\text{38^\circ}}.