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 \(\displaystyle m \angle C =m \angle D\).