In the given figure, AB and XY are diameters of a circle with center O. If ∠APX = 30°, find :

(i) ∠AOX

(ii) ∠APY

(iii) ∠BPY

(iv) ∠OAX

(i) Arc AX subtends ∠AOX at the center and ∠APX at the remaining part of the circle.

⇒ ∠AOX = 2∠APX (∵ angle subtended on center is twice the angle subtended on the remaining part of the circle.)

⇒ ∠AOX = 2 × 30° = 60°.

Hence, the value of ∠AOX = 60°.

(ii) From figure,

∠XPY = 90° [∵ angle in semicircle = 90°]

∴ ∠APY = ∠XPY - ∠APX = 90° - 30° = 60°.

Hence, the value of ∠APY = 60°.

(iii) From figure,

∠APB = 90° [∵ angle in semicircle = 90°]

∴ ∠BPY = ∠APB - ∠APY = 90° - 60° = 30°.

Hence, the value of ∠BPY = 30°.

(iv) In △AOX,

OA = OX (Radius of same circle)

∴ ∠OAX = ∠OXA

Since, sum of angles of a triangle = 180°

∴ ∠AOX + ∠OAX + ∠OXA = 180°

⇒ 60° + ∠OAX + ∠OAX = 180°

⇒ 2∠OAX = 120°

⇒ ∠OAX = $\dfrac{120°}{2}$

⇒ ∠OAX = 60°.

Hence, the value of ∠OAX = 60°.