In circle P with \( m \angle N P Q=96^{\circ} \) and \( N P=19 \) units, find the length of \( \overparen{N Q} \). Round to the nearest hundredth.

To find the length of \( \overline{NQ} \) in circle P given \( m \angle NPQ = 96^\circ \) and \( NP = 19 \) units, we can use the Law of Cosines. In triangle \( NPQ \): \[ NQ^2 = NP^2 + PQ^2 - 2 \cdot NP \cdot PQ \cdot \cos(m \angle NPQ) \] Since \( PQ \) is also the radius of the circle and equal to \( NP \), we have: \[ NQ^2 = NP^2 + NP^2 - 2 \cdot NP \cdot NP \cdot \cos(96^\circ) \] Therefore, \[ NQ^2 = 19^2 + 19^2 - 2 \cdot 19 \cdot 19 \cdot \cos(96^\circ) \] Calculating this step-by-step: 1. Calculate \( NP^2 \): \[ NP^2 = 19^2 = 361 \] 2. Use the cosine value: \(\cos(96^\circ) \approx -0.1045\). 3. Substitute values into the equation: \[ NQ^2 = 361 + 361 - 2 \cdot 19 \cdot 19 \cdot (-0.1045) \] \[ NQ^2 = 722 + 2 \cdot 361 \cdot 0.1045 \] \[ NQ^2 = 722 + 75.542 = 797.542 \] 4. Calculate \( NQ \): \[ NQ = \sqrt{797.542} \approx 28.22 \] Thus, the length \( \overline{NQ} \) is approximately \( 28.22 \) units rounded to the nearest hundredth.