In cyclic quadrilateral PQRS, P/3 = Q/5 = R/8 + 60. Find the largest angle in quadrilateral PQRS, in degrees.
By the angle bisector theorem, $PQ/PS = QR/SR$. Re-arranging, $\frac{QR}{SR} = \frac{5}{3} = \frac{PS}{PQ}$.
By the angle bisector theorem, $RS/RQ = PS/PQ$. Combining this with $\frac{QR}{SR} = \frac{PS}{PQ}$, we have $RS/RQ = QR/RP$, so triangle $PQR$ is isosceles, and $m\angle QPR = m\angle PQR = \theta$.
By the Law of Sines on triangle $RPQ$, $\frac{PQ}{\sin \theta} = \frac{RP}{\sin (180^\circ - 2 \theta)}$, or
\[\frac{PQ}{\sin \theta} = \frac{RP}{\sin 2 \theta} = 2 \cdot \frac{RP}{2 \sin \theta \cos \theta}.\]Then, $\frac{PQ}{2 \sin^2 \theta} = \frac{PR}{\cos \theta} = \frac{PS}{1 - \cos^2 \theta} = 3$. Hence,
\begin{align*}
PR &= 2 \sin^2 \theta, \\
PS &= 3 - 3 \cos^2 \theta.
\end{align*}(We know that $PS < PQ$, so $3 - 3 \cos^2 \theta < 2 \sin^2 \theta$.)
By the Law of Sines on triangle $QSR$, $\frac{RS}{\sin \theta} = \frac{QS}{\sin 2 \theta}$, so
\[RS = \frac{\sin \theta \cdot QS}{\sin 2 \theta} = \frac{\sin \theta \cdot \frac{5}{3} PQ}{2 \sin \theta \cos \theta} = \frac{5}{6} PQ \cot \theta.\]Then $SR = RQ + RS$, so
\[SR = PQ - PR + \frac{5}{6} PQ \cot \theta.\]Hence,
\[PQ - PR + \frac{5}{6} PQ \cot \theta = 8.\]From the equation $PR = 2 \sin^2 \theta,$ $\cot \theta = \frac{1 - \cos^2 \theta}{2 \sin \theta \cos \theta} = \frac{1 - \cos^2 \theta}{\sin 2 \theta}$.
Thus,
\[SR = PQ - 2 \sin^2 \theta + \frac{5}{6} PQ \cdot \frac{1 - \cos^2 \theta}{\sin 2 \theta}.\]Since $PQ = 3 \sin^2 \theta,$
\[SR = 3 \sin^2 \theta - 2 \sin^2 \theta + \frac{5}{6} \cdot 3 \sin^2 \theta \cdot \frac{1 - \cos^2 \theta}{\sin 2 \theta}.\]This reduces to
\[SR = \frac{19}{6} \sin^2 \theta + \frac{5}{6} \cdot \frac{1 - \cos^2 \theta}{\sin 2 \theta}.\]From the equation $3 - 3 \cos^2 \theta < 2 \sin^2 \theta$, $1 - \cos^2 \theta < 5 \sin^2 \theta$, so
\[SR < \frac{19}{6} \sin^2 \theta + \frac{5}{6} \cdot \frac{5 \sin^2 \theta}{\sin 2 \theta}.\]We can write $\sin 2 \theta$ in terms of $\sin \theta$ and $\cos \theta$: $\sin 2 \theta = 2 \sin \theta \cos \theta$. Hence,
\[SR < \frac{19}{6} \sin^2 \theta + \frac{25 \cos \theta}{6}.\]Recall that $PQ = 3 \sin^2 \theta = 2 - 2 \cos^2 \theta$. Then $2 \cos^2 \theta = 2 - PQ$, so $\cos \theta = \pm \sqrt{1 - \frac{PQ}{2}} = \pm \frac{\sqrt{5 - PQ}}{2}$.
Since $0^\circ < \theta < 180^\circ$, $0 < PQ < 4$. Therefore, $\cos \theta = \frac{\sqrt{5 - PQ}}{2}$.
Since $\sqrt{5 - PQ} > 0$, $SR$ is an increasing function of $\sin^2 \theta$, subject to $0 < PQ < 4$. Therefore, $SR$ is an increasing function of $\cos \theta$, or an increasing function of $\frac{\sqrt{5 - PQ}}{2}$.
Both $\frac{\sqrt{5 - PQ}}{2}$ and $\frac{\sqrt{5 - PQ}}{2}$ are positive. If $\frac{\sqrt{5 - PQ}}{2} > \cos \theta$, then
\[SR < \frac{19}{6} \sin^2 \theta + \frac{25 \sqrt{5 - PQ}}{6}.\]Otherwise, $SR \ge \frac{19}{6} \sin^2 \theta + \frac{25 \sqrt{5 - PQ}}{6}$.
\[
\begin{array}{c|c}
PQ & SR \\ \hline
0 & 6 \\
1 & \frac{25 \sqrt{4}}{6} = \frac{25}{3} \\
2 & \frac{19}{3} \cdot \frac{4}{9} + \frac{25 \sqrt{3}}{6} > \frac{25}{3} \\
3 & \frac{19}{3} \cdot \frac{9}{9} + \frac{25 \sqrt{2}}{6} > \frac{29}{3} \\
4 & \frac{19}{3} \cdot \frac{16}{9} + \frac{25 \sqrt{1}}{6} > \frac{32}{3}
\end{array}
\]We cannot have $SR$ greater than $\frac{32}{3}$, so $SR < \frac{32}{3}$.
Therefore, $SR < \frac{32}{3}$. But $PQSR$ is a quadrilateral inscribed in a circle, so $SR = RQ$. Hence, the largest angle is $\boxed{\frac{32}{3}}$.
To find the largest angle in cyclic quadrilateral PQRS, we can start by assigning variables to each angle. Let's call the angles P, Q, R, and S.
According to the given problem, we have the following ratios:
P/3 = Q/5 = R/8 + 60
Now, let's start solving for the values of P, Q, and R.
To find the value of P, we can multiply both sides of the equation P/3 = Q/5 by 3:
P = (Q/5) * 3
P = (3Q) / 5
Similarly, to find the value of R, we can multiply both sides of the equation R/8 + 60 = Q/5 by 8:
R + 480 = (8Q) / 5
R = (8Q) / 5 - 480
Now, we can substitute the values of P and R into the equation P + Q + R + S = 360° to find the value of S:
(3Q) / 5 + Q + (8Q) / 5 - 480 + S = 360°
To simplify this equation, we can find a common denominator:
(3Q + 5Q + 8Q - 2400 + 5S) / 5 = 360°
Adding like terms, we get:
16Q - 2400 + 5S = 1800°
Now, let's rearrange the equation to solve for S:
5S = 1800° - 16Q + 2400°
5S = 4200° - 16Q
S = (4200° - 16Q) / 5
Now, we have expressions for P, Q, R, and S, which are:
P = (3Q) / 5
R = (8Q) / 5 - 480
S = (4200° - 16Q) / 5
Since PQRS is a cyclic quadrilateral, the sum of opposite angles is equal to 180°. Therefore, the largest angle is either P or R.
To determine which angle is larger, we need to compare the expressions for P and R. Let's set them equal to each other and solve for Q:
(3Q) / 5 = (8Q) / 5 - 480
Multiplying both sides of the equation by 5 to eliminate the denominators, we get:
3Q = 8Q - 2400
Subtracting 8Q from both sides of the equation, we have:
-5Q = -2400
Dividing both sides by -5, we get:
Q = 480
Now, we can substitute the value of Q back into the expressions for P and R:
P = (3 * 480) / 5 = 144°
R = (8 * 480) / 5 - 480 = 768° - 480 = 288°
Since P (144°) is smaller than R (288°), the largest angle in the quadrilateral PQRS is angle R with a measure of 288 degrees.