Question

In the Given figure o is
the centre of the circle and
the Length an ARC AB is
TWICE the Length of ARC
BC.IF <AOB=104° FIND BY GIVING REASONS
<BOC.
<OAC.<BAC ​

Answer 1

∠BOC = 27∘ ∠OAC = 44.5∘

Step-by-step explanation:

Join AD and BD.

Arc B = 2 arc BC and ∠AOD = 180∘

∠BOC = ∠AOB = 1/2 * 108∘ = 54∘

Arc BC subtends ∠BOC and ∠CAB.

So we get ∠CAB = 1/2 ∠BOC = 1/2 * 54∘ = 27∘

In triangle AOC, ∠AOC = ∠AOB + ∠BOC = 54∘ + 27∘ = 81∘

AOC is an isoceles triangle as OA and OC are both radius of the circle.

So ∠OAC = 1/2 *(180∘ - ∠AOC) = 1/2 * (180 - 81) = 99/2 = 44.5∘