In the given figure, AB and CD are two equal chords of a circle, with center O.

If P is the mid-point of chord AB. Q is the mid-point of chord CD and ∠POQ = 150°, find ∠APQ.

We know that,

Equal chords are equidistant from the center.

∴ OP = OQ

In △ OPQ,

⇒ OP = OQ

⇒ ∠OQP = ∠OPQ = x (let) (Angles opposite to equal sides are equal)

By angle sum property of triangle,

⇒ ∠OQP + ∠OPQ + ∠POQ = 180°

⇒ x + x + 150° = 180°

⇒ 2x = 180° - 150°

⇒ 2x = 30°

⇒ x = $\dfrac{30°}{2}$ = 15°.

We know that,

A straight line drawn from the center of the circle to bisect the chord, which is not a diameter, is at right angles to the chord.

∴ ∠OPA = 90°

From figure,

⇒ ∠APQ = ∠OPA - ∠OPQ = 90° - x = 90° - 15° = 75°.

Hence, ∠APQ = 75°.