As shown in the figure, P is the first point on the circle, and the chord AB = root 3. PC is the bisector of angle APB. The angle BAC = 30 degrees 1. When PAC =? Degree, what is the maximum area of quadrilateral PACB? 2. When the angle PAC =? Degree, the quadrilateral PACB is trapezoidal? Explain the reason

As shown in the figure, P is the first point on the circle, and the chord AB = root 3. PC is the bisector of angle APB. The angle BAC = 30 degrees 1. When PAC =? Degree, what is the maximum area of quadrilateral PACB? 2. When the angle PAC =? Degree, the quadrilateral PACB is trapezoidal? Explain the reason

The maximum area of PACB is 9 ^ 3 / 4.2 when ∠ PAC = 120 ° and PACB is trapezoid. To make PACB trapezoid, arc AP = arc BC is needed. Because the circumferential angle of arc BC is 30 ° and that of arc AP is 30 °, so ∠ PAB = 90 ° and PC is the bisector of ∠ APB, arc AC = arc AC