If angle A is one of the base angles, then the other base angle must measure 50 degrees. Since 50 + 50 + x = 180 means x = 80, the vertex angle must measure 80 degrees.

If angle A is the vertex angle, the two base angles must be equal. Since 50 + x + x = 180 means x = 65, the two base angles must measure 65 degrees.

The only number given that is not possible is 95 degrees.

The sum of the measures of three angles of a triangle is $\displaystyle 180 ^{\circ }$, so we can set up the equation:

$\displaystyle \left ( x +30 \right )+\left ( 2x - 3\right ) + y = 180$

We can simplify and solve for $\displaystyle y$:

$\displaystyle 3x +27 + y = 180$

$\displaystyle 3x +27 + y-3x-27 = 180-3x-27$

$\displaystyle y = -3x + 153$

The sum of the measures of the angles of a triangle is $\displaystyle 180^{\circ }$, so simplify and solve for $\displaystyle y$ in the equation:

$\displaystyle x + (2x-30) + y = 180$

$\displaystyle (x + 2x)-30 + y = 180$

$\displaystyle y + 3x-30 = 180$

$\displaystyle y + 3x-30 -3x + 30= 180 -3x + 30$

$\displaystyle y = 210 -3x$

The measures of the angles of a triangle total $\displaystyle 180^{\circ }$, so if two angles measure $\displaystyle 44^{\circ }$ and we call $\displaystyle x$ the measure of the third, then 

$\displaystyle x + 44 + 44 = 180$

$\displaystyle x + 88 = 180$

$\displaystyle x + 88 - 88 = 180- 88$

$\displaystyle x= 92^{\circ } > 90^{\circ }$

This makes the triangle obtuse.

Also, since the triangle has two congruent angles (the $\displaystyle 44^{\circ }$ angles), the triangle is also isosceles.

We invoke the SAS Inequality Theorem, which states that, given two triangles $\displaystyle \Delta ABC$ and $\displaystyle \Delta DEF$, with $\displaystyle AB = DE, BC = EF$, $\displaystyle m\angle B < m \angle E$ ( the included angles), then $\displaystyle AC < DF$ - that is, the side opposite the greater angle has the greater length. Since $\displaystyle \angle B$ is an acute angle, and $\displaystyle \angle E$ is a right angle, we have just this situation. This makes (b) the greater.

(a) The measures of the angles of a linear pair total 180, so:

$\displaystyle x+ y = 180$

$\displaystyle x+ 89 = 180$

$\displaystyle x+ 89 - 89 = 180- 89$

$\displaystyle x = 91$

(b) The Triangle Exterior-Angle Theorem states that the measure of an exterior angle is equal to the sum of its remote interior angles. Therefore, $\displaystyle y = 54 + 35 = 89$.

Therefore (a) is the greater quantity.

The two angles at bottom are marked as congruent. Each of these two angles forms a linear pair with a $\displaystyle 140 ^{\circ }$ angle, so it is supplementary to that angle, making its measure $\displaystyle (180-140)^{\circ } = 40^{\circ }$.  Therefore, the other marked angle also measures $\displaystyle 40^{\circ }$.

The sum of the measures of the interior angles of a triangle is $\displaystyle 180^{\circ }$, so:

$\displaystyle y + 40 + 40 = 180$

$\displaystyle y + 80 = 180$

$\displaystyle y + 80 -80 = 180 -80$

$\displaystyle y = 100$

The quantities are equal.

The Triangle Exterior-Angle Theorem states that the measure of an exterior angle is equal to the sum of its remote interior angles. Therefore, 

$\displaystyle x = 78 + 62 = 140$, 

making the quantities equal.

$\displaystyle \Delta ABC$ is equilateral, so

$\displaystyle AB = AC =20, BC = 20$.

In $\displaystyle \Delta CBD$, we are given that

$\displaystyle BC = CD = 20, BD = 25$.

Since the triangles have two pair of congruent sides, the third side with the greater length is opposite the angle of greater measure. Therefore, 

$\displaystyle m \angle BCD >m \angle BAC$.

Since $\displaystyle \angle BAC$ is an angle of an equilateral triangle, its measure is $\displaystyle 60^{\circ }$, so $\displaystyle m \angle BCD >60^{\circ }$.

Corresponding angles of similar triangles are congruent, so, since $\displaystyle \bigtriangleup ABC \sim \bigtriangleup DEF$, it follows that 

$\displaystyle m \angle D = m \angle A = 60 ^{\circ }$

By similarity, $\displaystyle m \angle B = m \angle E$ and $\displaystyle m \angle C = m \angle F$, and we are given that $\displaystyle m \angle E = m \angle F$, so 

$\displaystyle m \angle B = m \angle C$

Also,

$\displaystyle m \angle A + m \angle B + m \angle C = 180^{\circ }$

$\displaystyle 60^{\circ }+ m \angle C + m \angle C = 180^{\circ }$

$\displaystyle 60^{\circ }+2 \cdot m \angle C = 180^{\circ }$

$\displaystyle 2 \cdot m \angle C = 120^{\circ }$

$\displaystyle m \angle C = 60^{\circ }$,

and $\displaystyle m \angle C =m \angle D$.