Greatest possible measure of $\angle A$ in an isosceles triangle $ABC$ I am a high schooler studying for the SAT, and I came across this question.

In isosceles triangle $ABC$, side $\overline{BC}$ is longer than the other two sides. If the degree measure of $\angle A$ is a multiple of 22, calculate the greatest possible measure of $\angle C$.
(A) 78
(B) 88
(C) 75
(D) 86
(E) 79

The correct answer according to the study guide is (E) 79.
But in the very first sentence of the question, it is stated that side $BC$ is longer than the other two sides, and we know that sides $BA$ and $AC$ are equal in length because this is an isosceles triangle.
That means that $\angle A > 60°$.
Were they just throwing in the part about $BC$ being longer to confuse me?
 A: 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
A: 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.
A: 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 $79^\circ$.
A: I actually ran into one these type of questions on the real SAT years ago when I took it.  The way it was explained to me and my fellow students taking the test was that these questions are generated for the test.  They have a set sentence structure and words are pulled from a list to fill in the blanks.  In this question we can easily identity the problem.  "Side $BC$ is longer than the other two sides"  The word longer is the problem  here.  If you were to change the word "longer" to "shorter" the question would be correct and the answer $E$ $79^\circ$ would also be the correct answer.  You get this by knowing that angles $b$ and $c$ are equal and $a$ is a multiple of $22$.  All three angles must add up to $180^\circ$ in all.
That means $c+b+(22a)=180^\circ$ but $c$ and $b$ are the same value so $2b+22a = 180^\circ$ which can be rearranged to $b=\frac{180^\circ-22a}{2}$.
Now some may say this equation cannot be solved easily but that is not true.  The question is asking for the greatest measurement that angle '$c$' can be.  That means that the angle $a$ is as small as it possibly can be.  Thus we plug in $1$ for '$a$' giving us $22\times 1$ which is $22$.  We use $1$ because $1 \times 22^\circ$ is the smallest possible angle that is a multiple of $22$.  We can then solve a rather simple equation of $a b = \frac{180^\circ-22^\circ}{2} = 79^\circ$ the answer.
So in conclusion this question was auto generated and simply pulled one wrong word from a list.  This is a kind of a large problem that hasn't really been addressed for $10+$ years.
