1. **Problem Statement:** We are given a circle with diameter $\overline{AB}$ and points $C$ and $D$ on the circle. The angle between radii $PA$ and $PD$ is $(7x+1)^\circ$, the angle between radii $PB$ and $PC$ is $(9x-7)^\circ$, and the angle between radii $PD$ and $PC$ is a right angle (90°). We need to find the measure of the major arc $\overset{\frown}{ACD}$ in degrees. 2. **Key Concepts:** - The center $P$ of the circle is the vertex of the central angles. - The sum of central angles around point $P$ is 360°. - The major arc $\overset{\frown}{ACD}$ corresponds to the central angle $\angle APB$ minus the minor arc $\overset{\frown}{AB}$. 3. **Set up the equation:** The central angles around $P$ are: $$\angle APD = (7x+1)^\circ$$ $$\angle BPC = (9x-7)^\circ$$ $$\angle DPC = 90^\circ$$ Since $\angle APD + \angle DPC + \angle BPC = 180^\circ$ (because $\overline{AB}$ is a diameter and these three angles form a straight line), we write: $$ (7x+1) + 90 + (9x-7) = 180 $$ 4. **Simplify and solve for $x$:** $$ 7x + 1 + 90 + 9x - 7 = 180 $$ $$ 16x + 84 = 180 $$ $$ 16x = 180 - 84 $$ $$ 16x = 96 $$ $$ x = \frac{96}{16} = 6 $$ 5. **Calculate each angle:** $$ \angle APD = 7(6) + 1 = 42 + 1 = 43^\circ $$ $$ \angle BPC = 9(6) - 7 = 54 - 7 = 47^\circ $$ 6. **Find the measure of minor arc $\overset{\frown}{ACD}$:** The minor arc $\overset{\frown}{ACD}$ corresponds to the sum of $\angle APD + \angle DPC + \angle BPC = 43 + 90 + 47 = 180^\circ$. 7. **Find the measure of major arc $\overset{\frown}{ACD}$:** Since $\overline{AB}$ is a diameter, the entire circle is 360°, so the major arc $\overset{\frown}{ACD}$ is: $$ 360^\circ - 180^\circ = 180^\circ $$ **Final answer:** The arc measure of major arc $\overset{\frown}{ACD}$ is $180^\circ$.