1999 AMC 8 Problem 21

Use supplementary angles at the two intersections

Then use a triangle angle sum near the $40^\circ$ angle

Label the vertices as below.

We then have that $$\angle ABC = 180^{\circ} - 100^{\circ} = 80^{\circ}$$ by supplementary angles. Then we have $$\angle CED = 180^{\circ} - 110^{\circ} = 70^{\circ}$$ again by supplementary angles. Using the sum of the interior angles of a triangle is $180^{\circ},$ we get $$\begin{aligned} \angle ECD &= 180^{\circ} - 70^{\circ} - 40^{\circ} \\ &= 70^{\circ}. \end{aligned}$$ Then, using vertical angles, we have $$\angle ACB = \angle ECD = 70^{\circ}.$$ Finally, using the sum of the interior angles of a triangle, we get $$\angle A = 180^{\circ} - 80^{\circ} - 70^{\circ} = 30^{\circ}.$$

Thus, B is the correct answer.