Question

exercise #3: in the diagram below, points e, f, and i are collinear.
(a) what does the fact that e, f, and i are collinear tell you about the measure of ∠efi?
(b) name two pairs of supplementary angles.
(c) name one pair of complementary angles.
exercise #4: it is given that ∠a and ∠b are complementary. if the measure of ∠b is 15 more than twice the measure of ∠a, determine the measure of both angles algebraically.
exercise #5: it is given that m∠bae = 4x + 11 and m∠qrp = 6x - 1. if ∠bae and ∠qrp are supplementary, determine how many degrees larger is the measure of ∠qrp than the measure of ∠bae?
exercise #6: in the diagram below, (overline{rt}) and (overline{uv}) intersect at point s.
(a) name two angles that are both supplementary to ∠ust.
(b) if m∠ust = 130°, then find m∠usr and m∠tsv.
(c) what do you think is true about angles that are supplementary to the same angle?

Explanation:

Exercise #3

(a)

Since points $E$, $F$, and $I$ are collinear, $\angle EFI = 180^{\circ}$ (a straight - line angle measures $180^{\circ}$).

(b)

Supplementary angles add up to $180^{\circ}$. Two pairs of supplementary angles could be $\angle EFG$ and $\angle GFI$, $\angle EFH$ and $\angle HFI$ (assuming $G$ and $H$ are as shown in the diagram).

(c)

Complementary angles add up to $90^{\circ}$. If there is a right - angle formed in the diagram (not clearly shown in the description but if say $\angle GFH$ is a right - angle), then $\angle GFH$ and $\angle HFI$ (if $\angle GFI$ is a straight line) could be a pair of complementary angles.

Exercise #4

Let the measure of $\angle A=x$. Then the measure of $\angle B = 2x + 15$.
Since $\angle A$ and $\angle B$ are complementary, $\angle A+\angle B=90^{\circ}$.
So, $x+(2x + 15)=90$.

Step1: Combine like terms

$3x+15 = 90$.

Step2: Subtract 15 from both sides

$3x=90 - 15=75$.

Step3: Divide both sides by 3

$x=\frac{75}{3}=25$.
So, $m\angle A = 25^{\circ}$ and $m\angle B=2\times25 + 15=50 + 15 = 65^{\circ}$.

Exercise #5

Since $\angle BAE$ and $\angle QRP$ are supplementary, $m\angle BAE+m\angle QRP = 180^{\circ}$.
So, $(4x + 11)+(6x-1)=180$.

Step1: Combine like terms

$10x+10 = 180$.

Step2: Subtract 10 from both sides

$10x=180 - 10 = 170$.

Step3: Divide both sides by 10

$x = 17$.
$m\angle BAE=4x + 11=4\times17+11=68 + 11=79^{\circ}$.
$m\angle QRP=6x-1=6\times17-1=102 - 1 = 101^{\circ}$.
The difference is $m\angle QRP-m\angle BAE=101 - 79 = 22^{\circ}$.

Exercise #6

(a)

$\angle USR$ and $\angle TSV$ are both supplementary to $\angle UST$ because $\angle UST+\angle USR = 180^{\circ}$ and $\angle UST+\angle TSV=180^{\circ}$ (linear pairs of angles).

(b)

If $m\angle UST = 130^{\circ}$, since $\angle UST+\angle USR = 180^{\circ}$, then $m\angle USR=180 - 130=50^{\circ}$.
Also, $\angle UST$ and $\angle TSV$ are vertical angles (opposite angles formed by the intersection of two lines), so $m\angle TSV=m\angle USR = 50^{\circ}$.

(c)

Angles that are supplementary to the same angle are congruent. Let $\angle A$ be an angle and $\angle B$ and $\angle C$ be two angles such that $\angle A+\angle B = 180^{\circ}$ and $\angle A+\angle C = 180^{\circ}$. Then $\angle B=\angle C$.

Answer:

Exercise #3
(a) $m\angle EFI = 180^{\circ}$
(b) $\angle EFG$ and $\angle GFI$, $\angle EFH$ and $\angle HFI$
(c) Example: $\angle GFH$ and $\angle HFI$ (if applicable)
Exercise #4
$m\angle A = 25^{\circ}$, $m\angle B = 65^{\circ}$
Exercise #5
$22^{\circ}$
Exercise #6
(a) $\angle USR$, $\angle TSV$
(b) $m\angle USR = 50^{\circ}$, $m\angle TSV = 50^{\circ}$
(c) Angles supplementary to the same angle are congruent.