In the figure (ii) given below, AP and BP are tangents to the circle with centre O. If ∠CBP = 25° and ∠CAP = 40°, find

(i) ∠ADB

(ii) ∠AOB

(iii) ∠ACB

(iv) ∠APB.

(i) ∠CDB = ∠CBP (∵ angles in alternate segments are equal.)

∴ ∠CDB = 25° ….(i)

Similarly, ∠CDA = ∠CAP = 40° (∵ angles in alternate segments are equal.)

∴ ∠ADB = ∠CDA + ∠CDB = 40° + 25° = 65°.

Hence, the value of ∠ADB = 65°.

(ii) Arc AB subtends ∠AOB at the centre and ∠ADB at the remaining part of the circle.

⇒ ∠AOB = 2∠ADB (As angle subtended on centre is twice the angle subtended on the remaining part of the circle).

⇒ ∠AOB = 2 × 65°
⇒ ∠AOB = 130°.

Hence, the value of ∠AOB = 130°.

(iii) ACBD is a cyclic quadrilateral.

∴ ∠ACB + ∠ADB = 180° (∵ sum of opposite angles = 180°.)

⇒ ∠ACB + 65° = 180°
⇒ ∠ACB = 180° - 65° = 115°.

Hence, the value of ∠ACB = 115°.

(iv) From figure,

⇒ ∠AOB + ∠APB = 180°
⇒ 130° + ∠APB = 180°
⇒ ∠APB = 180° - 130° = 50°.

Hence, the value of ∠APB = 50°.