As shown in the figure, the center O of the regular triangle ABC is exactly the center of the sector ode, and the point B is in the sector. No matter how the sector ode rotates around the point O, the area of the overlap between △ ABC and the sector is equal to 13% of the area of △ ABC. What is the central angle of the sector? Explain your reasons

As shown in the figure, the center O of the regular triangle ABC is exactly the center of the sector ode, and the point B is in the sector. No matter how the sector ode rotates around the point O, the area of the overlap between △ ABC and the sector is equal to 13% of the area of △ ABC. What is the central angle of the sector? Explain your reasons

When the central angle of a sector is 120 degrees, the area of the overlapping part of △ ABC and sector is always equal to 13% of the area of △ ABC. It is proved that: (1) when the central angle of a sector coincides with the central angle of an equilateral triangle, obviously, the area of the overlapping part of △ ABC and sector is equal to 13% of the area of △ ABC; (2) when the central angle of a sector coincides with the central angle of an equilateral triangle, the area of △ ABC and sector is equal to 13% of the area of △ ABC

As shown in the figure, it is known that PA and Pb cut ⊙ o at two points a and B, connecting AB, and the lengths of PA and Pb are two of the equations x2-2mx + 3 = 0, ab = M. try to find: (1) the radius of ⊙ o; (2) the area of the figure (that is, the shadow part) enclosed by PA, Pb and ab

(1) Connecting OA, ob, ∵ PA = Pb, (1 point) ∵ △ = (- 2m) 2-4 × 3 = 0, ∵ M2 = 3, m ＞ 0, ∵ M = 3, ∵ x2-23x + 3 = 0, ∵ X1 = x2 = 3, ∵ PA = Pb = AB = 3, ∵ ABP equilateral triangle, ∵ APB = 60 °, (3 points) ∵ apo = 30 °,