Question

In the given figure, DC is a chord of a circle with centre O. DC is produced to B such that BC = OC and ∠OBD = 30°. If AOB is straight line, then ∠AOD = ?

Answer 1

Answer:

AB is a chord of a circle with centre O AB is produced to C such that BO = BC

CO is joined and produced to meet the circle at D

We shall prove x°=3y°

We have

BC=OB

∠OCB=∠BOC=y°

[Angles opposite to equal sides are equal ]

∠OBA=∠BOC+∠OCB

[Ext angle of a △ is equal to the sum of the opposite interior angles ]

⇒∠OBC=y° +y°=2y°

OA=OB...[Radii of the same circle ]

∠OAB=∠OBA....[Angles opp. To equal sides of a △]

=2y

∠AOD=∠OAC+∠OCA

=2y°+y°=3y°

[Exterior angle - Sum of opposite interrior angles]

⇒x° =3y°

Answer 2

Step-by-step explanation:

OC = OB ( given)

∠ CBO = ∠COB

∠B = ∠ O

30° = 30°

∠B + ∠C + ∠O = 180

30 + ∠C + 30 = 180

∠C = 180 -60

∠BCO = 120

∠ BCO + ∠OCD = 180

∠OCD = 180 - 120

∠ OCD = 60

OC = OD ( radii )

∠OCD = ∠ODC

∠OCD + ∠ODC +∠COD = 180

60 + 60 +∠ COD = 180

120 + ∠ COD = 180

∠COD = 60

∠COB + ∠COD + ∠DOA = 180

30 + 60 + ∠DOA = 180

∠DOA OR AOD = 90