A quadrilateral PQRS is inscribed in a circle of centre O, such that PQ is a diameter and . Find the value of .

Correct — D. Both angle expressions sit in the stem as images: the given is angle PSR = 120° and the quantity asked for is angle QPR.

PQRS is cyclic, so opposite angles are supplementary.
angle PQR = 180° − 120° = 60°

PQ is a diameter and R lies on the circle, so the angle in a semicircle applies.
angle PRQ = 90°

The three angles of triangle PQR sum to 180°.
angle QPR = 180° − 90° − 60° = 30° → option (d).

Why the others are wrong

• (a)
  50° — 50° does not survive the check: with the right angle fixed at R, angle PQR would be 40°, forcing angle PSR to 140° rather than the 120° the stem gives.
• (b)
  60° — 60° is angle PQR, the value the supplement hands you on the way. Taken as the answer it forces angle PSR to 150°, so it is one line too early.
• (c)
  40° — 40° would leave angle PQR at 50°, which makes angle PSR 130°. The right angle at R is fixed by the diameter and cannot be traded away to absorb the difference.

Two circle theorems settle this, and each supplies one angle of triangle PQR.

Opposite angles of a cyclic quadrilateral are supplementary, so angle PSR and angle PQR add to 180°. That turns the 120° given at S into 60° at Q.

An angle in a semicircle is a right angle: because PQ is a diameter, any point R on the circle sees PQ at 90°. With two angles of the triangle known, the third follows from the 180° sum.

Key facts

• Opposite angles of a cyclic quadrilateral sum to 180°.
• An angle subtended by a diameter at any point on the circle is 90°, by Thales' theorem.
• The angles of a triangle sum to 180°, so two known angles fix the third.
• Here angle PQR = 60°, angle PRQ = 90° and angle QPR = 30°.

Common traps

• Copying 120° onto angle PQR instead of taking its supplement 60°
• Placing the right angle at P or Q rather than at R, the vertex not on the diameter
• Answering with the intermediate 60° when the question asks for angle QPR

No directly related past PYQ was found.