Answer:
d. 67.2°
Explanation:
Since the leg dimensions of the larger triangle ACB are given, it is convenient to use those to determine the angle.
In this geometry, all of the right triangles are similar. We note that ∠ABD is the larger of the two acute angles in the triangle. In ΔACB, the corresponding angle is ∠ACB. With respect to that angle, we are given the lengths of adjacent an opposite legs of the triangle, so the tangent relation is useful:
Tan = Opposite/Adjacent
tan(∠ABD) = tan(∠ADB) = BA/BC = 19/8
∠ABD = arctan(19/8) ≈ 67.166°
∠ABD ≈ 67.2°