B)
as you recall yesterday, the "far arc near arc" formula, well, notice, the angle at XDY intercepts the circle at XY, with a "near arc" of 100°, and a "far arc" of 360° - 100° = 260°.
so using the far arc near arc formula
[tex]\bf \measuredangle D=\cfrac{260-100}{2}\implies \measuredangle D=\cfrac{160}{2}\implies \measuredangle D = 80[/tex]
the twin angles at A and C will take the slack from all interior angles of 180° - 80° = 100°, so then ∡A = ∡C = 50°.
