OA, OB, and OC are all radii of the circle so they have the same length, 12 in. ∆ AOB is therefore isosceles, so the angles at the vertices A and B are congruent. The interior angles of any triangle sum to 180° in measure, so
30° + 2 m∠A = 180° ⇒ m∠A = 75°
Draw an altitude in ∆ AOB that connects the vertex O to the midpoint of AB; this is the height h of ∆ AOB and bisects ∆ AOB into two 15°-75°-90° right triangles. Solve for h and the length of the base AB using trigonometry:
sin(15°) = (AB/2) / (12 in) ⇒ AB = (24 in) sin(15°) ≈ 6.212 in
sin(75°) = h / (12 in) ⇒ h = (12 in) sin(75°) ≈ 11.591 in
Then the area of ∆ AOB is
1/2 (AB) h = (144 in²) sin(15°) sin(75°)
= (72 in²) (cos(60°) - cos(90°))
= 36 in²
where we use the identity
sin(x) sin(y) = 1/2 (cos(x - y) - cos(x + y))
Let m∠BOC = θ. Then
(area of sector BOC) / θ = (area of ⊙O) / 360°
Solve for θ :
θ = (area of sector BOC) / (area of ⊙O) • 360°
θ = (36 in²) / (π (12 in)²) • 360°
θ = (90/π)° ≈ 28.648°