In the figure, O is the centre of the circle and the length of arc AB is twice the length of arc BC. If angle AOB = 108°, find:
(i) ∠CAB (ii) ∠ADB
(i) Join AD and DB Arc B = 2 arc BC and ∠AOB = 180° ∴ ∠BOC = 1 ∠AOB = `1/2 xx108°` = 54° Now, Arc BC subtends ∠BOC at the centre and ∠CAB at the remaining part of the circle. ∴ `∠CAB = 1/2 ∠BOC` =` 1/2xx 54°` = 27°
(ii) Again, Arc AB subtends ∠AOB at the centre and ∠ACB at the remaining part of the circle. `∠ ACB = 1/2 ∠AOB` =` 1/2 xx 108°` = 54 In cyclic quadrilateral ADBC ∠ADB + ∠ACB =180° [sum of opposite angles] ⇒ ∠ADB + 54° = 180° ⇒ ∠ADB = 180° - 54° ⇒ ∠ADB = 126°
Solution In the Figure, O is the Centre of the Circle and the Length of Arc Ab is Twice the Length of Arc Bc. If Angle Aob = 108°, Find: Concept: Arc and Chord Properties - the Angle that an Arc of a Circle Subtends at the Center is Double that Which It Subtends at Any Point on the Remaining Part of the Circle.