The measure of an exterior angle of a triangle is the sum of the measures of its remote interior angles, so 

$\displaystyle w = x + y = 72 + 43 = 115 ^{\circ }$

In an isosceles triangle, the base angles are the same. Also, the three angles of a triangle add up to $\displaystyle 180^{\circ}$.

So, subtract the vertex angle from $\displaystyle 180^{\circ}$. You get $\displaystyle 116^{\circ}$.

Because there are two base angles you divide $\displaystyle 116^{\circ}$ by $\displaystyle 2$, and you get $\displaystyle 58^{\circ}$.

The measure of an exterior angle of a triangle is the sum of the measures of its remote interior angles, so 

$\displaystyle \angle 3 = \angle 1 + \angle 2$.

We also have the following constraints:

$\displaystyle 45^{\circ } \leq m \angle 1 \leq 50^{\circ }$

$\displaystyle 60^{\circ } \leq m \angle 2 \leq 70 ^{\circ }$

Then, by the addition property of inequalities,

$\displaystyle 45^{\circ } + 60 ^{\circ }\leq m \angle 1 +m \angle 2 \leq 50^{\circ } + 70 ^{\circ }$

$\displaystyle 105 ^{\circ }\leq m \angle 3 \leq 120^{\circ }$

Therefore, the measure of $\displaystyle \angle 3$ must fall in that range. Of the given choices, only $\displaystyle 110^{\circ }$ falls in that range.

The measure of an exterior angle of a triangle is the sum of the measures of its remote interior angles, so 

$\displaystyle \angle 3 = \angle 1 + \angle 2$

or 

$\displaystyle \angle 2 = \angle 3 - \angle 1$

Therefore, the maximum value of $\displaystyle \angle 2$ is the least possible value of $\displaystyle \angle 1$ subtracted from the greatest possible value of $\displaystyle \angle 3$:

$\displaystyle 140^{\circ } - 45^{\circ } = 95 ^{\circ }$

The minimum value of $\displaystyle \angle 2$ is the greatest possible value of $\displaystyle \angle 1$ subtracted from the least possible value of $\displaystyle \angle 3$:

$\displaystyle 130^{\circ } - 50^{\circ } = 80 ^{\circ }$

Therefore, 

$\displaystyle 80^{\circ } \leq \angle 2 \leq 95^{\circ }$

Since all of the choices fall in this range, all are possible measures of $\displaystyle \angle 2$.

All the angles in a triangle must add up to $\displaystyle 180$.

$\displaystyle 32+46+y=180$

$\displaystyle 78+y=180$

$\displaystyle y=102$

All the angles in a triangle must add up to $\displaystyle 180.$

$\displaystyle a+35+35=180$

$\displaystyle a+70=180$

$\displaystyle a=110$

All the angles in a triangle must add up to $\displaystyle 180$.

$\displaystyle b+28+37=180$

$\displaystyle b+65=180$

$\displaystyle b=115$

By the Isosceles Triangle Theorem, an isosceles triangle has two congruent interior angles. There are two possible scenarios if one angle has measure $\displaystyle 70^{\circ }$:

Scenario 1: The other two angles are congruent to each other. The degree measures of the interior angles of a triangle total $\displaystyle 180^{\circ }$, so if we let $\displaystyle x$ be the common measure of those angles:

$\displaystyle x + x + 70 = 180$

$\displaystyle 2x+70 = 180$

$\displaystyle 2x= 110$

$\displaystyle x = 55$

This makes (b) a possible answer.

Scenario 2: One of the other angles measures $\displaystyle 70^{\circ }$ also, making (c) a possible answer. The degree measure of the third angle is

$\displaystyle 180 - (70 + 70) = 180 - 140 = 40$,

making (a) a possible answer. Therefore, the correct choice is (a), (b), or (c).

A scalene triangle has three sides of different measure, so, by way of the Converse of the Isosceles Triangle Theorem, each angle is of different measure as well. We can therefore eliminate $\displaystyle 54^{\circ }$ immediately. 

Also, if the triangle also has a $\displaystyle 72^{\circ }$ angle, then, since the total of the degree measures of the angles is $\displaystyle 180^{\circ }$, it follows that the third angle has measure

$\displaystyle 180 - (54+72) = 180 - 126 = 54 ^{\circ }$.

Therefore, the triangle has two angles that measure the same, and $\displaystyle 72^{\circ }$ can be eliminated.

Similarly, if the triangle also has a $\displaystyle 63^{\circ }$ angle, then, since the total of the degree measures of the angles is $\displaystyle 180^{\circ }$, it follows that the third angle has measure

$\displaystyle 180 - (54+63) = 180 - 117= 63 ^{\circ }$.

The triangle has two angles that measure $\displaystyle 63^{\circ }$. This choice can be eliminated.

$\displaystyle 126^{\circ }$ can be eliminated, since the third angle would have measure

$\displaystyle 180 - (54+126) = 180 - 180= 0^{\circ }$,

an impossible situation since angle measures must be positive.

The remaining possibility is $\displaystyle 108^{\circ }$. This would mean that the third angle has measure

$\displaystyle 180 - (54+108) = 180 - 162= 18^{\circ }$.

The three angles have different measures, so the triangle is scalene. $\displaystyle 108^{\circ }$ is the correct choice.

Above is the figure described.

The measures of the interior angles of a triangle total $\displaystyle 180^{\circ }$, so the measure of $\displaystyle \angle ACB$ is

$\displaystyle m \angle ACB = 180^{\circ } -( m \angle A + m \angle B)$

$\displaystyle = 180^{\circ } -( 20^{\circ } + 32^{\circ })$

$\displaystyle = 180^{\circ } - 52^{\circ }$

$\displaystyle = 128^{\circ }$

Since $\displaystyle \overrightarrow{CD}$ bisects this angle, 

$\displaystyle m \angle BCD= \frac{1}{2}m \angle ACB = \frac{1}{2} \cdot 128^{\circ } =64^{\circ }$

and 

$\displaystyle m \angle CDB = 180^{\circ } -( m \angle BCD + m \angle B)$

$\displaystyle = 180^{\circ } -( 64 ^{\circ } + 32^{\circ } )$

$\displaystyle = 180^{\circ } -96^{\circ }$

$\displaystyle = 84^{\circ }$