1. **Problem Statement:** Given circle T with points P, Q, R, and S on the circumference, and \(\angle PTQ \cong \angle RTS\), find the length of chord \(PQ\). 2. **Known Information:** - \(TR = 3\) units - \(TS = 4\) units - \(\angle RTS = 66^\circ\) - \(\angle PTQ = \angle RTS = 66^\circ\) 3. **Key Concept:** In a circle, equal central angles subtend chords of equal length. Since \(\angle PTQ = \angle RTS\), chords \(PQ\) and \(RS\) are equal in length. 4. **Find length of chord \(RS\):** Use the Law of Cosines in triangle \(TRS\): $$RS^2 = TR^2 + TS^2 - 2 \cdot TR \cdot TS \cdot \cos(\angle RTS)$$ Substitute values: $$RS^2 = 3^2 + 4^2 - 2 \cdot 3 \cdot 4 \cdot \cos(66^\circ)$$ Calculate: $$RS^2 = 9 + 16 - 24 \cdot \cos(66^\circ) = 25 - 24 \cdot 0.4067 = 25 - 9.76 = 15.24$$ 5. **Calculate \(RS\):** $$RS = \sqrt{15.24} \approx 3.9$$ 6. **Conclusion:** Since \(PQ = RS\), the length of \(PQ\) is approximately 4 units. **Final answer:** \(\boxed{4}\) units.