Extend $\displaystyle \overline{AB}$ as seen in the figure below:

The measure of an exterior angle of a triangle is equal to the sum of the measures of its remote interior angles; specifically,

$\displaystyle m \angle CAD = m \angle B + m \angle C$,

and

 $\displaystyle m \angle CAD =( x+ y) ^{\circ }$

However, $\displaystyle m \angle CAD > z ^{\circ }$, so, by substitution,

$\displaystyle x+y>z$

Below is the referenced triangle along with $\displaystyle \bigtriangleup DEF$, an equilateral triangle with sides of length 10:

As an angle of an equilateral triangle, $\displaystyle \angle D$ has measure $\displaystyle 60 ^{\circ }$. Applying the Side-Side-Side Inequality Theorem, since $\displaystyle AB = DE$, $\displaystyle AC = DF$, and $\displaystyle BC > EF$, it follows that $\displaystyle m \angle A > m \angle D$, so $\displaystyle m \angle A > 60^{\circ }$.

Also, since $\displaystyle AB = BC$, by the Isosceles Triangle Theorem, $\displaystyle m \angle B = m \angle C$. Since $\displaystyle m \angle A > 60^{\circ }$, and the sum of the measures of the angles of a triangle is $\displaystyle 180 ^{\circ }$, it follows that

$\displaystyle m \angle B + m \angle C < 120^{\circ }$

Substituting and solving:

$\displaystyle m \angle B + m \angle B < 120^{\circ }$

$\displaystyle 2 m \angle B < 120^{\circ }$

$\displaystyle 2 m \angle B \div 2 < 120^{\circ } \div 2$

$\displaystyle m \angle B < 60^{\circ }$.

Examine the diagram below, in which two triangles matching the given descriptions have been superimposed.

Note that $\displaystyle AB = DE$ and $\displaystyle m \angle B = m \angle E$. The two question marks need to be replaced by $\displaystyle C$ and $\displaystyle F$. No matter how you place these two points, $\displaystyle AC = DF$. However, with one replacement, $\displaystyle BC > EF$; with the other replacement, $\displaystyle BC < EF$. Therefore, the information given is insufficient to answer the question.

$\displaystyle AC = DF, AB = DE, BC = EF$, so, by the Side-Side-Side Principle, since there are three pairs of congruent corresponding sides between the triangles, we can say they are congruent - that is,

$\displaystyle \Delta ABC \cong \Delta DEF$.

Corresponding angles of congruent sides are congruent, so $\displaystyle m \angle A = m \angle D$.

A scalene triangle, by definition, has sides all of different lengths. Since all of the given choices fit that criterion, the correct choice is that all can be scalene.

$\displaystyle m \angle B = 90^{\circ}$

$\displaystyle m \angle C = 50 ^{\circ }$

The sum of the measures of the angles of a triangle is 180, so

$\displaystyle m \angle A + m \angle B + m \angle C = 180$

$\displaystyle m \angle A +90 + 50 = 180$

$\displaystyle m \angle A +140= 180$

$\displaystyle m \angle A +140-140= 180-140$

$\displaystyle m \angle A = 40^{\circ }$

$\displaystyle m \angle C > m \angle A$, so the side opposite $\displaystyle \angle C$, which is $\displaystyle \overline{AB}$, is longer than the side opposite $\displaystyle \angle A$, which is $\displaystyle \overline{BC}$. This makes (a) the greater quantity.

To compare the lengths of $\displaystyle \overline{AB}$ and $\displaystyle \overline{BC}$ from the angle measures, it is necessary to know which of their opposite angles - $\displaystyle \angleC$$\displaystyle \angle C$ and $\displaystyle \angle A$, respectively - is the greater angle. Since $\displaystyle \angle A$ is the obtuse angle, it has the greater measure, and $\displaystyle \overline{BC}$ is the longer side. This makes (b) greater.

Since $\displaystyle \angle B$ is the obtuse angle of $\displaystyle \Delta ABC$, 

$\displaystyle \left ( AC \right )^{2} >\left ( AB \right )^{2} + \left ( BC \right )^{2} = 7^{2} + 24^{2} = 49 + 576 = 625$.

$\displaystyle \left ( AC \right )^{2} > 625$,

$\displaystyle AC > \sqrt{625}$

$\displaystyle AC >25$,

so (a) is the greater quantity.

Use the Triangle Inequality:

$\displaystyle AC < AB + BC$

$\displaystyle AC < 5 + 12$

$\displaystyle AC < 17$

This makes (b) the greater quantity.

By the Converse of the Pythagorean Theorem, 

$\displaystyle AC = \sqrt{(AB)^{2}+(BC)^{2}}= \sqrt{5^{2}+12^{2}}= \sqrt{25+144}=\sqrt{169} = 13$

if and only if $\displaystyle \angle B$ is a right angle. 

However, if $\displaystyle \angle B$ is acute, then $\displaystyle AC < 13$;  if $\displaystyle \angle B$ is obtuse, then $\displaystyle AC > 13$.

Since we do not know whether $\displaystyle \angle B$ is acute, right, or obtuse, we cannot determine whether (a) or (b) is greater.