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 \( \overparen{N Q} \) in circle P, we can use the Law of Cosines. We know the angle \( m \angle N P Q = 96^\circ \) and the radius \( N P = 19 \) units. Since \( N P \) and \( P Q \) are both radii of the circle, they are equal. Let's denote the length of \( N Q \) as \( c \). The sides of the triangle are: - \( a = N P = 19 \) units - \( b = P Q = 19 \) units - \( C = \angle N P Q = 96^\circ \) Using the Law of Cosines: \[ c^2 = a^2 + b^2 - 2ab \cdot \cos(C) \] Substituting in our values: \[ c^2 = 19^2 + 19^2 - 2 \cdot 19 \cdot 19 \cdot \cos(96^\circ) \] Calculating step-by-step: \[ c^2 = 361 + 361 - 2 \cdot 361 \cdot \cos(96^\circ) \] Now, the approximate value of \( \cos(96^\circ) \) is \( -0.1045 \): \[ c^2 = 361 + 361 - 2 \cdot 361 \cdot (-0.1045) \] \[ = 722 + 75.5089 \approx 797.5089 \] Taking the square root to find \( c \): \[ c = \sqrt{797.5089} \approx 28.24 \] Thus, the length of \( \overparen{N Q} \) is approximately **28.24** units.