1999 AMC 8 Problems/Problem 21

[asy] unitsize(12); draw((0,0)--(20,0)--(1,-10)--(9,5)--(18,-8)--cycle); draw(arc((1,-10),(1+19/sqrt(461),-10+10/sqrt(461)),(25/17,-155/17),CCW)); draw(arc((19/3,0),(19/3-8/17,-15/17),(22/3,0),CCW)); draw(arc((900/83,-400/83),(900/83+19/sqrt(461),-400/83+10/sqrt(461)),(900/83 - 9/sqrt(97),-400/83 + 4/sqrt(97)),CCW)); label(rotate(30)*" $40^\circ$ ",(2,-8.9),ENE); label(" $100^\circ$ ",(21/3,-2/3),SE); label(" $110^\circ$ ",(900/83,-317/83),NNW); label(" $A$ ",(0,0),NW); label(" $B$ ", (20,0), NW);[/asy] Note that $\angle B=180-100-40=40^\circ$ . So $\angle A=180-110-40=\boxed{30^\circ}$ .

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1999 AMC 8 Problems/Problem 21