In cyclic quadrilatera \(PQRS,\)
\(\frac{\angle P}{2} = \frac{\angle Q}{3} = \frac{\angle R}{4}.\) Find the largest angle of quadrilateral PQRS, in degrees.
The sum of the opposite angles in a cyclic quadrilateral sum to 180 degrees.
Thus, p=2/3q, q=q, r=4/3q
2/3q+4/3q=180
6/3q=180
q=90
p=60
r=120
s=90
And, the largest angle is angle r, which measures 120 degrees.
