Question

In the given figure, AB = BC = DC and ∠AOB = 50°.
(i) ∠AOC
(ii) ∠AOD
(iii) ∠BOD
(iv) ∠OAC
(v) ∠ODA

Answer 1

Since arc AB and BC are equal.
So, ∠AOB = ∠BOC = 50°
Now,
∠AOC = ∠AOB + ∠BOC = 50° + 50° = 100°
As arc AB, arc BC and arc CD so, 
∠AOB = ∠BOC = ∠COD = 50°
∠AOD = ∠AOB + ∠BOC + ∠COD = 50° + 50° + 50° = 150°
Now, ∠BOD = ∠BOC + ∠COD 
∠BOD = 50° + 50° 
∠BOD = 100°

The triangle thus formed, ΔAOC is an isosceles triangle with OA = OC as they are radii of the same circle.
 Thus ∠OAC = ∠OCA as they are opposite angles of equal sides of an isosceles triangle.
The sum of all the angles of a triangle is 180°
So, ∠AOC + ∠OAC + ∠OCA = 180°
2∠OAC + 100° = 180°     as, ∠OAC = ∠OCA
2∠OAC = 180° - 100° 
2∠OAC = 80°
∠OAC = 40°

as ∠OCA = ∠OAC So,
∠OCA = ∠OAC = 40°

The triangle thus formed, ΔAOD is an isosceles triangle with OA = OD as they are radii of the same circle.
Thus, ∠OAD = ∠ODA as they are opposite angles of equal sides of an isosceles triangle.
The sum of all the angles of a triangle is 180°
So, ∠AOD + ∠OAD + ∠ODA = 180°
2∠OAD + 150° = 180°         as, ∠OAD = ∠ODA
2∠OAD = 180° - 150° 
∠OAD = 30°

as ∠OAD = ∠ODA    So,
∠OAD = ∠ODA  = 15°.