From the center of the O circle, draw the radii OB and OS.
The diameter of the circle, by condition, is 12 cm, BD = 12 cm, then R = ОВ = OC = ВD / 2 = 12/2 = 6 cm.
The BOS triangle is isosceles, since OB = OS = R = 6 cm.
Let’s draw the height OH, which is also the bisector and median of the triangle.
Then, in a right-angled triangle BOH, the hypotenuse ОВ = 6 cm, the leg BН = ВС / 2 = 6/2 = 3 cm.
Determine the size of the acute angle BOH. SinBОН = BН / ОB = 3/6 = 1/2.
Angle BOC = arcsin (1/2) = 30. Then the central angle of BOC is equal to: BOC = 2 * BOC = 2 * 30 = 60.
The degree measure of the ВС arc is equal to the central angle of the ВС and is equal to 60.
The inscribed angle BAC also rests on the BC arc and is equal to half of its degree measure.
Angle BAC = BOC / 2 = 60/2 = 30.
Answer: The BAC angle is 30.