Greatest possible measure of $\angle A$ in an isosceles triangle $ABC$

Your claim that $\triangle ABC$ is obtuse is mistaken. Suppose $\angle A$ were a right angle, for example. Then you would have an isosceles right triangle, and $BC$, the hypotenuse, would certainly be the longest side.

Observe that $BC$ will be equal to the other two sides if the triangle is equilateral—that is when $\angle A=60^\circ$. So $BC$ will be the longest side whenever $\angle A$ is larger than $60^\circ$.

But it seems to me that the problem is still insoluble: the correct answer has $\angle A = 66^\circ$ and so $\angle C = 57^\circ$, which is not a choice.

The proposed solution of $\angle C = 79^\circ$ is clearly wrong. This makes $\angle A = 22^\circ$, and then $BC$ is not the longest side as stated.

It seems to me that you are right! I think the book got things confused.

Indeed If $BC$ has the longest length, angle A should be the largest, with the other $2$ being smaller (and equal to each other as the triangle is isosceles).

However, the argument you wrote down has a significant error. Just because $BC$ is the longest side, it is not implied that the triangle is obtuse. Consider a triangle with a $70^\circ$ and two $55^\circ$ angles. This triangle is clearly isosceles with two equal angles and yet, not obtuse as all angles are less than $90^\circ$.

I think what they meant is that $BC$ is the smallest side. This will allow for $A$ to be $22^\circ$ and the other two angles to be $79^\circ$.

If BC is longer than the other 2 sides, then the other 2 sides must be equal because the triangle is isosceles. If indeed they claim that the answer is E, that implies that angle C (79°) is bigger than angle A and thus side AB must be bigger than side BC. So this is clearly incorrect. This is not the first time I have come across such a question from ACT, SAT