In the figure (i) given below, AB and XY are diameters of a circle with centre O. If ∠APX = 30°, find

(i) ∠AOX

(ii) ∠APY

(iii) ∠BPY

(iv) ∠OAX.

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

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

⇒ ∠AOX = 2 × 30°
⇒ ∠AOX = 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 the same circle.)

∴ ∠OAX = ∠OXA

Since sum of angles of a triangle = 180°

∴ ∠AOX + ∠OAX + ∠OXA = 180°
⇒ 60° + ∠OAX + ∠OAX = 180°
⇒ 2∠OAX = 180° - 60°
⇒ 2∠OAX = 120°
⇒ ∠OAX = 60°.

Hence, the value of ∠OAX = 60°.