To find the measure of minor arc \(TU\) in the circle centered at \(O\), we begin by analyzing the given information regarding the tangents and angles.
Given:
- \(PT\) and \(PU\) are tangents to the circle at points \(T\) and \(U\) respectively.
- Angle \(TPO\) measures \(33^\circ\).
From the properties of tangents to a circle, we know that the angle formed between the tangent and the radius drawn to the point of tangency is \(90^\circ\). Therefore, we can establish:
\[ \angle OTP = 90^\circ \quad \text{and} \quad \angle OUP = 90^\circ \]
Next, consider the triangle \(TOP\). Using the angles we have:
- \(\angle TPO = 33^\circ\)
- Hence, \(\angle OTP = 90^\circ\)
Using the triangle angle sum property:
\[ \angle TOP + \angle OTP + \angle TPO = 180^\circ \]
Substituting the known angles:
\[ \angle TOP + 90^\circ + 33^\circ = 180^\circ \]
This simplifies to:
\[ \angle TOP + 123^\circ = 180^\circ \]
Therefore:
\[ \angle TOP = 180^\circ - 123^\circ = 57^\circ \]
Now, because angles \(\angle TON\) and \(\angle UON\) each measure \(57^\circ\) (as they are base angles of the isosceles triangle formed by the equal tangents \(PT\) and \(PU\)), we can relate this angle to the minor arc \(TU\).
The central angle \(TU\) corresponding to minor arc \(TU\) is twice the angle \(TOP\):
\[ \text{Measure of arc } TU = 2 \times \angle TOP = 2 \times 57^\circ = 114^\circ \]
Thus, the measure of minor arc \(TU\) is:
\[ \boxed{114} \] degrees.
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