PQRS is a cyclic quadrilateral and PQ is the diameter of the circle. If ∠RPQ = 38°, then what is the value (in degrees) of ∠PSR?
128
Step 1 — Angle in semicircle: Since PQ is the diameter, \(\angle PRQ=90^\circ\).
Step 2 — Find \(\angle PQR\) in \(\triangle PQR\):
\[\angle PQR=180^\circ-90^\circ-38^\circ=52^\circ\]
Step 3 — Cyclic quadrilateral property: Opposite angles sum to \(180^\circ\), so \(\angle PSR+\angle PQR=180^\circ\).
\[\angle PSR=180^\circ-52^\circ=128^\circ\]
Therefore \(\angle PSR=128^\circ\).
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