Supplementary angles and complementary angles have measures totaling $\displaystyle 180^{\circ }$ and $\displaystyle 90^{\circ }$, respectively.

(a) The measure of an angle complementary to a $\displaystyle 55 ^{\circ }$ angle is $\displaystyle 90^{\circ } - 55^{\circ } = 35^{\circ }$

(b) The measure of an angle supplementary to a $\displaystyle 135 ^{\circ }$ angle is $\displaystyle 180^{\circ } - 135 ^{\circ } = 45^{\circ }$

This makes (b) greater.

Two angles are complementary if their degree measures total 90. Therefore, 

$\displaystyle m \angle 1 + m \angle 2 = 90$

Since $\displaystyle m \angle 1 =2 m \angle 2 - 50$, we can substitute, and we can solve for $\displaystyle m\angle2$:

$\displaystyle (2 m \angle 2 - 50)+ m \angle 2 = 90$

$\displaystyle 3 m \angle 2 - 50 = 90$

$\displaystyle 3 m \angle 2 - 50 + 50 = 90 + 50$

$\displaystyle 3 m \angle 2 = 140$

$\displaystyle 3 m \angle 2 \div 3 = 140 \div 3$

$\displaystyle m \angle 2 = 46 \frac{2}{3}$

$\displaystyle m \angle 1 = 90 - m \angle 2 = 90 - 46 \frac{2}{3} = 43 \frac{1}{3}$

$\displaystyle m \angle 2 > m \angle 1$, making (B) the greater quantity.

$\displaystyle \angle 1$ and $\displaystyle \angle 2$ form a linear pair, so 

$\displaystyle m \angle1 + m \angle 2 = 180$

Since $\displaystyle m \angle1 = m \angle 2 + 30$, this can be rewritten as

$\displaystyle m \angle1 - 30 = m \angle 2$, and the first equation can be rewritten as:

$\displaystyle m \angle1 + m \angle1 - 30 = 180$

$\displaystyle 2 m \angle1 - 30 = 180$

$\displaystyle 2 m \angle1 - 30 + 30 = 180 + 30$

$\displaystyle 2 m \angle1 = 210$

$\displaystyle 2 m \angle1 \div 2 = 210 \div 2$

$\displaystyle m \angle1 = 105$

The two angles form a linear pair and therefore their measures total $\displaystyle 180^{\circ }$. We check all of the pairs for this sum.

$\displaystyle 55+35 = 90$

$\displaystyle 64+ 36 = 100$

$\displaystyle 112+ 68 = 180$

$\displaystyle 124+76 = 200$

The correct pair is $\displaystyle 112^{\circ }, 68^{\circ }$.

The measure of an exterior angle of a triangle, which here is $\displaystyle \angle 3$, is the sum of the measures of its remote interior angles, which here are  $\displaystyle \angle 1$ and $\displaystyle \angle 2$. Therefore, we are looking for the sum of the first two angle measures to be equal to the third.

$\displaystyle 42 ^{\circ }+ 72 ^{\circ } = 114 ^{\circ }$

$\displaystyle 30 ^{\circ } + 80 ^{\circ } = 110 ^{\circ }$

$\displaystyle 56 ^{\circ } + 86 ^{\circ } = 142 ^{\circ }$

$\displaystyle 25 ^{\circ }+ 88 ^{\circ } = 113 ^{\circ }$

All four triples satisfy this condition.

The degree measure of $\displaystyle \angle 1$ is five degrees greater than twice that of $\displaystyle \angle 2$ - that is, 

$\displaystyle m \angle 1 = 2 m \angle 2 + 5$

Since $\displaystyle \angle 1$ and $\displaystyle \angle 2$ form a linear pair, 

 $\displaystyle m\angle1 + m\angle 2 = 180$, and by substitution, 

$\displaystyle ( 2 m \angle 2 + 5)+ m\angle 2 = 180$

$\displaystyle 3 m \angle 2 + 5 = 180$

$\displaystyle 3 m \angle 2 + 5 - 5 = 180 - 5$

$\displaystyle 3 m \angle 2 = 175$

$\displaystyle 3 m \angle 2 \div 3 = 175\div 3$

$\displaystyle m \angle 2 = 58 \frac{1}{3} > 55$

This makes (A) greater.

$\displaystyle \angle 1$ and $\displaystyle \angle 2$ form a linear pair and therfore, the the sum of their degree measures is $\displaystyle 180^{\circ }$. 

$\displaystyle m\angle 1 + m\angle 2 = 180^{\circ }$

$\displaystyle m\angle 1 + 48 ^{\circ } = 180^{\circ }$

$\displaystyle m\angle 1 + 48 ^{\circ } - 48 ^{\circ } = 180^{\circ } - 48 ^{\circ }$

$\displaystyle m\angle 1 = 132 ^{\circ }$

$\displaystyle 3 m \angle 1 = 2 m \angle 2$, so 

$\displaystyle 3 \left (m \angle 1 \right ) \div 2 = 2\left ( m \angle 2 \right )\div 2$

$\displaystyle \frac{3}{2} m \angle 1 = m \angle 2$

The measure of an exterior angle of a triangle, which here is $\displaystyle \angle 3$, is the sum of the measures of its remote interior angles, which here are  $\displaystyle \angle 1$ and $\displaystyle \angle 2$.

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

$\displaystyle m \angle 1 + \frac{3}{2} m \angle 1 = 125 ^{\circ }$

$\displaystyle \frac{5}{2} m \angle 1 = 125 ^{\circ }$

$\displaystyle \frac{2}{5} \cdot \frac{5}{2} m \angle 1 = \frac{2}{5} \cdot 125 ^{\circ }$

$\displaystyle m \angle 1 =50 ^{\circ }$

The equation for solving for the slope of a line is $\displaystyle m=\frac{y2-y1}{x2-x1}$. 

Thus, if $\displaystyle (x1,y1)=(10, 25)$ and $\displaystyle (x2,y2)=(5,10)$, then:

$\displaystyle m=\frac{25-10}{10-5}=\frac{15}{5}=3$

The equation for solving for the slope of a line is $\displaystyle m=\frac{y2-y1}{x2-x1}$. 

Thus, if $\displaystyle (x1,y1)=(4.5, 2.5)$ and $\displaystyle (x2,y2)=(3.5,6.5)$, then:

$\displaystyle m=\frac{6.5-2.5}{3.5-4.5}=\frac{4}{-1}=-4$