Question

In the given figure, O is the center of the circle and the length of arc AB is twice the length of arc BC. If ∠AOB = 100°,
find: (i) ∠BOC (ii) ∠OAC

Answer 1

We know that when two arcs are in ratio 2: 1 then the subtended by them is also in ratio 2: 1
As given arc AB is twice the length of arc BC.
Therefore, arc AB: arc BC = 2: 1
Hence, ∠AOB: ∠BOC = 2: 1

Now given that ∠AOB = 100°.
So, ∠BOC = `1/2∠"AOB" = 1/2 xx 100° = 50°`

Now, ∠AOC = ∠AOB + ∠BOC = 100° + 50° = 150°.
The triangle thus formed, ∠AOC 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, ∠COA + ∠OAC + ∠OCA = 180°
2∠OAC + 150° = 180° as, ∠OAC = ∠OCA 
2∠OAC = 180° - 150°
2∠OAC = 30°
∠OAC = 15°
as ∠OCA = ∠OAC So,
∠OCA = ∠OAC = 15°.