Compare Angles A and B in an Obtuse Triangle: Vertex Angle Analysis

To solve this problem, we need to compare $\angle B$ and $\angle A$ in an obtuse triangle ABC.

A triangle is classified as obtuse when one of its angles is greater than $90^\circ$. In such a triangle, the largest angle is the obtuse angle.

Without loss of generality, if we consider any angle of the triangle, say $\angle C$, to be the obtuse angle, it must be that $\angle C > 90^\circ$. This makes $\angle C$ the largest angle.

Given the angle sum property of triangles ($\angle A + \angle B + \angle C = 180^\circ$), the sum of the two non-obtuse angles ($\angle A$ and $\angle B$) must be less than $90^\circ$, hence ensuring $\angle C$ remains the largest.

Since $\angle B$ and $\angle A$ must both be less than $90^\circ$, and the problem requires determining which is larger without any specific constraints on $\angle C$, we observe:

• If $\angle C$ is indeed obtuse, then $\angle A$ and $\angle B$ must add up to less than $90^\circ$, leading to $\angle B$ generally being greater than $\angle A$ under typical conditions unless otherwise specified.
• This result denotes that $\angle B$ being comparably larger than $\angle A$ unless specified otherwise by additional conditions, which are absent here.

Therefore, $\angle B > \angle A$.

Hence, in the context of the problem's provided choices and lacking other conditions, the solution is $\angle B > \angle A$.

Thus, the larger angle is $\angle B > \angle A$.