Question: For items 25 to 30, refer to the image below. Graph shape and math: a circle on the right-center with points N at the top, Q at the upper-right, R at the lower-right, and S at the lower-left on the circumference; M is outside the circle on the left. Segments are drawn from M to N and M to R, with MN tangent/meeting the circle at N, MR crossing the circle through P and ending at R, and inside the circle the chords N-S, N-Q, Q-R, and S-R are connected. position_hint=center If m\angle PR = 115^\circ, what is the measure of arc PNR? (Just type the numerical value.) In the figure, quadrilateral NQRS is inscribed in the circle. If m\angle NQR = 105^\circ, what is the measure of m\angle NSR? (Just type the numerical value.) Let line segment MN be tangent to the circle at point N, and let MR be a secant line that intersects the circle at points P and R. If the length of the tangent segment MN=12 and the inner chord segment PR=10, what is the length of the external secant segment MP? (Just type the numerical value.)

1. **Problem 1:** Given $m\angle PR = 115^\circ$, find the measure of arc $PNR$. - The angle $\angle PR$ is an inscribed angle that intercepts arc $PNR$. - The measure of an inscribed angle is half the measure of its intercepted arc. 2. **Formula:** $$m\angle = \frac{1}{2} m\text{(intercepted arc)}$$ 3. **Calculation:** $$115 = \frac{1}{2} m(arc\ PNR)$$ Multiply both sides by 2: $$2 \times 115 = m(arc\ PNR)$$ $$230 = m(arc\ PNR)$$ 4. **Answer:** The measure of arc $PNR$ is $230$ degrees. --- 1. **Problem 2:** Quadrilateral $NQRS$ is inscribed in the circle. Given $m\angle NQR = 105^\circ$, find $m\angle NSR$. - Opposite angles of an inscribed quadrilateral sum to $180^\circ$. 2. **Formula:** $$m\angle NQR + m\angle NSR = 180^\circ$$ 3. **Calculation:** $$105 + m\angle NSR = 180$$ Subtract 105 from both sides: $$m\angle NSR = 180 - 105$$ $$m\angle NSR = 75$$ 4. **Answer:** The measure of $m\angle NSR$ is $75$ degrees. --- 1. **Problem 3:** Given tangent segment $MN = 12$, chord segment $PR = 10$, find external secant segment $MP$. - The tangent-secant theorem states: $$MN^2 = MP \times MR$$ - Since $MR = MP + PR$, we have: $$12^2 = MP \times (MP + 10)$$ 2. **Calculation:** $$144 = MP^2 + 10 MP$$ Rearranged: $$MP^2 + 10 MP - 144 = 0$$ 3. **Solve quadratic:** $$MP = \frac{-10 \pm \sqrt{10^2 - 4 \times 1 \times (-144)}}{2} = \frac{-10 \pm \sqrt{100 + 576}}{2} = \frac{-10 \pm \sqrt{676}}{2}$$ $$MP = \frac{-10 \pm 26}{2}$$ 4. **Possible solutions:** $$MP = \frac{-10 + 26}{2} = \frac{16}{2} = 8$$ $$MP = \frac{-10 - 26}{2} = \frac{-36}{2} = -18$$ 5. **Answer:** Length cannot be negative, so $MP = 8$. --- **Final answers:** - Measure of arc $PNR$ is $230$. - Measure of $m\angle NSR$ is $75$. - Length of external secant segment $MP$ is $8$.