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 

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

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

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

$\displaystyle = 84^{\circ }$

If $\displaystyle \bigtriangleup BCD$ is isosceles, then by the Isosceles Triangle Theorem, two of its angles must be congruent. 

Case 1: $\displaystyle m \angle B = m \angle BCD= 74^{\circ }$

Since $\displaystyle \overrightarrow{CD}$ bisects $\displaystyle \angleACB$$\displaystyle \angle ACB$ into two congruent angles, one of which must be $\displaystyle m \angle BCD$, 

$\displaystyle m \angle ACB = 2 \cdot m \angle BCD = 2 \cdot 74^{\circ } = 148^{\circ }$

However, this is impossible, since $\displaystyle \angle ACB$ and $\displaystyle \angle B$ are two angles of the original triangle; their total measure is

 $\displaystyle m \angle ACB+ m \angle B = 148^{\circ }+ 74 ^{\circ } = 222 ^{\circ } > 180 ^{\circ }$

Case 2: $\displaystyle m \angle B = m \angle BDC= 74^{\circ }$

Then, since the degree measures of the interior angles of a triangle total $\displaystyle 180^{\circ }$,

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

$\displaystyle = 180^{\circ } - (74^{\circ } + 74^{\circ } )$

$\displaystyle = 180 ^{\circ } - 148^{\circ }$

$\displaystyle = 32^{\circ }$

Since $\displaystyle \overrightarrow{CD}$ bisects $\displaystyle \angleACB$$\displaystyle \angle ACB$ into two congruent angles, one of which must be $\displaystyle m \angle BCD$, 

$\displaystyle m \angle ACB = 2 \cdot m \angle BCD = 2 \cdot 32^{\circ } = 64^{\circ }$

and

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

$\displaystyle = 180^{\circ } - (74^{\circ } + 64^{\circ } )$

$\displaystyle = 180^{\circ } - 138^{\circ }$

$\displaystyle = 42^{\circ }$

Case 3: $\displaystyle m \angle BDC= m \angle BCD$

Then

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

$\displaystyle 74= 180^{\circ } -(m \angle BCD+ m \angle BCD)$

$\displaystyle 74= 180^{\circ } -2 \cdot m \angle BCD$

$\displaystyle 2 \cdot m \angle BCD = 106^{\circ }$

$\displaystyle m \angle BCD = 53^{\circ }$

$\displaystyle m \angle ACB = 2 \cdot m \angle BCD = 2 \cdot 53^{\circ } = 106^{\circ }$

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

$\displaystyle = 180^{\circ } - ( 74^{\circ }+ 106^{\circ } )$

$\displaystyle = 0^{\circ }$, which is not possible.

Therefore, the only possible measure of $\displaystyle \angle A$ is $\displaystyle 42^{\circ }$.

The degree measures of the interior angles of a triangle total 180 degrees, so 

$\displaystyle x + (x+26) + (2x+ 1) = 180$

$\displaystyle x +x+ 2x + 26 + 1 = 180$

$\displaystyle 4x+27 = 180$

$\displaystyle 4x+27 - 27 = 180 - 27$

$\displaystyle 4x = 153$

$\displaystyle \frac{4x}{4} = \frac{153}{4}$

$\displaystyle x = 38 \frac{1}{4}$

One angle measures $\displaystyle 38\frac{1}{4}^{\circ }$

The other two angles measure 

$\displaystyle x+ 26 = 38\frac{1}{4} + 26 = 64\frac{1}{4} ^{\circ }$

and 

$\displaystyle 2x+ 1 = 2 \left\(38\frac{1}{4}\right$+ 1 = 77 \frac{1}{2} ^{\circ }\).

We want the greatest of the three, or $\displaystyle 77 \frac{1}{2} ^{\circ }$.

The figure referenced is below:

Since $\displaystyle \bigtriangleup ADC$ is an isosceles right triangle, its acute angles - in particular, $\displaystyle \angle ADC$ - measure $\displaystyle 45^{\circ }$ each. Since this angle forms a linear pair with $\displaystyle \angle CDB$:

$\displaystyle m \angle CDB = 180^{\circ } - m \angle ADC = 180^{\circ } - 45 ^{\circ } = 135^{\circ }$.

 $\displaystyle \bigtriangleup BDC$ is also isosceles, so, by the Isosceles Triangle Theorem, it has two congruent angles. Since $\displaystyle \angle CDB$ is obtuse, and no triangle has two obtuse angles:

$\displaystyle m \angle BCD = m \angle B$.

Also, $\displaystyle \angle ADC$ is an exterior angle of $\displaystyle \bigtriangleup BDC$, whose measure is equal to the sum of those of its two remote interior angles, which are the congruent angles $\displaystyle m \angle BCD = m \angle B$. Therefore,

$\displaystyle m \angle ADC = m \angle BCD + m \angle B$

$\displaystyle 45^{\circ }= m \angle B + m \angle B$

$\displaystyle 45^{\circ }=2 m \angle B$

$\displaystyle m \angle B = 22\frac{1}{2}^{\circ }$