Answer:
B. 29º
Step-by-step explanation:
Given that BCDE is a parallelogram, then [tex]BE \parallel CD[/tex] and A, E and D are collinear. Then, angle E inside triangle ABE and angle D inside parallelogram BCDE have the same measure. That is:
[tex]\angle D = \angle E = 51^{\circ}[/tex] (1)
In addition, the sum of internal angles in triangles equals 180º, which implies that:
[tex]\angle A + \angle B + \angle E = 180^{\circ}[/tex] (2)
If we know that [tex]\angle B = 100^{\circ}[/tex] and [tex]\angle E = 51^{\circ}[/tex], then [tex]\angle A[/tex] is:
[tex]\angle A = 180^{\circ}-\angle B - \angle E[/tex]
[tex]\angle A = 180^{\circ}-100^{\circ}-51^{\circ}[/tex]
[tex]\angle A = 29^{\circ}[/tex]
Hence, correct answer is B.
