13. if $ \overleftrightarrow{de} \parallel \overleftrightarrow{gf} $, what is $ m\angle dfg $?
$ m\angle dfg = \underline{\quad\quad}

14.
match each angle pair with its description.
angle 1 and angle 2
angle 2 and angle 5
angle 4 and angle 5
vertical angles
linear pair
complementary angles

15.
if \( \angle a $ and $ \angle b $ are complementary and $ m\angle a = 67^\circ $, what is $ m\angle b $?
$ m\angle b = \underline{\quad\quad}^\circ

16.
what is \( m\angle nrp $?

Question

13. if $ \overleftrightarrow{de} \parallel \overleftrightarrow{gf} $, what is $ m\angle dfg $?
$ m\angle dfg = \underline{\quad\quad}

14.
match each angle pair with its description.
angle 1 and angle 2
angle 2 and angle 5
angle 4 and angle 5
vertical angles
linear pair
complementary angles

15.
if \( \angle a $ and $ \angle b $ are complementary and $ m\angle a = 67^\circ $, what is $ m\angle b $?
$ m\angle b = \underline{\quad\quad}^\circ

16.
what is \( m\angle nrp $?

Accepted Answer

### Question 13
# Explanation:
## Step1: Identify the relationship
Since $ \overleftrightarrow{DE} \parallel \overleftrightarrow{GF} $, the angle $ 75^\circ $ and $ \angle DFG $ are same - side interior angles? Wait, no, actually, the angle at $ D $ ( $ 75^\circ $) and $ \angle DFG $ are same - side interior angles? Wait, no, let's look at the transversal. The angle $ \angle EDF = 75^\circ $, and $ \overleftrightarrow{DE}\parallel\overleftrightarrow{GF} $, so $ \angle DFG $ and $ \angle EDF $ are same - side interior angles? Wait, no, actually, the sum of same - side interior angles is $ 180^\circ $. Wait, no, maybe it's a consecutive interior angle. Wait, the angle at $ D $ ( $ 75^\circ $) and $ \angle DFG $: since $ DE\parallel GF $, and the transversal is the line through $ D $ and $ F $, then $ \angle EDF $ and $ \angle DFG $ are same - side interior angles, so $ m\angle DFG=180 - 75=105^\circ $? Wait, no, maybe I made a mistake. Wait, the angle given is $ 75^\circ $ at $ D $, between $ DE $ and the transversal. Then, since $ DE\parallel GF $, the angle $ \angle DFG $ and $ \angle EDF $ are same - side interior angles, so they are supplementary. So $ m\angle DFG = 180^{\circ}-75^{\circ}=105^{\circ} $? Wait, no, maybe it's a corresponding angle? No, corresponding angles are equal. Wait, maybe the angle at $ D $ is $ 75^\circ $, and $ \angle DFG $ is supplementary to it because they are same - side interior angles. So $ 180 - 75 = 105 $.
## Step2: Calculate the measure
$ m\angle DFG=180^{\circ}-75^{\circ}=105^{\circ} $
# Answer:
$ 105^\circ $

### Question 14
# Brief Explanations:
- **Angle 1 and Angle 2**: A linear pair of angles is formed when two lines intersect. Angle 1 and Angle 2 are adjacent and their non - common sides form a straight line, so they are a Linear Pair.
- **Angle 2 and Angle 5**: Complementary angles are two angles whose sum is $ 90^{\circ} $. From the diagram, Angle 1 is a right angle (since there is a right - angle symbol), so Angle 2+Angle 5 = 90^{\circ}, so they are Complementary Angles.
- **Angle 4 and Angle 5**: Vertical angles are formed by two intersecting lines and are opposite each other. Angle 4 and Angle 5 are not vertical angles. Wait, no, Angle 1 and Angle 3 are vertical angles, Angle 2 and Angle 4? Wait, no, let's look again. The lines: one horizontal, one vertical, and one slant. Angle 4 and Angle 5: since Angle 1 is $ 90^{\circ} $, Angle 4 + Angle 5=90^{\circ}? No, wait, Angle 1 is a right angle, so Angle 4 + Angle 5 = 90^{\circ}? No, maybe I messed up. Wait, the vertical angles: Angle 1 and Angle 3 are vertical angles, Angle 2 and Angle 4 are vertical angles? Wait, no, the diagram: the horizontal line, vertical line, and a slant line. Angle 1 and Angle 3: vertical angles. Angle 2 and Angle 4: vertical angles. Then Angle 4 and Angle 5: since Angle 1 is $ 90^{\circ} $, Angle 4 + Angle 5 = 90^{\circ}, so they are Complementary? No, the problem says to match:

Angle 1 and Angle 2: Linear Pair (they form a straight line, adjacent, sum to $ 180^{\circ} $)

Angle 2 and Angle 5: Complementary Angles (sum to $ 90^{\circ} $ because Angle 1 is $ 90^{\circ} $)

Angle 4 and Angle 5: Wait, no, maybe Angle 4 and Angle 5: no, maybe Angle 1 and Angle 3 are vertical angles. Wait, the options are Vertical Angles, Linear Pair, Complementary Angles.

So:

Angle 1 and Angle 2: Linear Pair

Angle 2 and Angle 5: Complementary Angles

Angle 4 and Angle 5: Wait, no, maybe Angle 4 and Angle 5: no, maybe Angle 1 and Angle 3 are vertical angles. Wait, the problem's angle pairs:

Angle 1 and Angle 2: Linear Pair (adjacent, form a straight line)

Angle 2 and Angle 5: Complementary Angles (sum to $ 90^{\circ} $)

Angle 4 and Angle 5: Complementary Angles? No, maybe I made a mistake. Wait, the right - angle symbol is at Angle 1, so Angle 1 is $ 90^{\circ} $. So Angle 4 + Angle 5 = 90^{\circ}, so they are Complementary. Angle 2 and Angle 5: if Angle 1 is $ 90^{\circ} $, and Angle 2 + Angle 5 + Angle 1? No, maybe the diagram is such that Angle 2 and Angle 5: no, let's re - express:

- Linear Pair: two adjacent angles that form a straight line (sum to $ 180^{\circ} $): Angle 1 and Angle 2 (since they are adjacent and their non - common sides form a straight line)
- Complementary Angles: sum to $ 90^{\circ} $: Angle 2 and Angle 5 (if Angle 1 is $ 90^{\circ} $, then Angle 2 + Angle 5 = 90^{\circ}), Angle 4 and Angle 5 (Angle 4 + Angle 5 = 90^{\circ})
- Vertical Angles: opposite angles formed by intersecting lines: Angle 1 and Angle 3, Angle 2 and Angle 4

But the problem gives three angle pairs: Angle 1 and Angle 2, Angle 2 and Angle 5, Angle 4 and Angle 5, and three descriptions: Vertical Angles, Linear Pair, Complementary Angles.

So:

Angle 1 and Angle 2: Linear Pair

Angle 2 and Angle 5: Complementary Angles

Angle 4 and Angle 5: Complementary Angles? No, maybe the problem has a typo, but according to the diagram, let's assume:

Angle 1 and Angle 2: Linear Pair

Angle 2 and Angle 5: Complementary Angles

Angle 4 and Angle 5: Complementary Angles (but that can't be). Wait, maybe Angle 4 and Angle 5: no, maybe Angle 1 and Angle 3 are vertical angles, but the problem's angle pairs don't include Angle 1 and Angle 3. So perhaps:

Angle 1 and Angle 2: Linear Pair

Angle 2 and Angle 5: Complementary Angles

Angle 4 and Angle 5: Complementary Angles (but the descriptions are Vertical Angles, Linear Pair, Complementary Angles). So maybe:

Angle 1 and Angle 2: Linear Pair

Angle 2 and Angle 5: Complementary Angles

Angle 4 and Angle 5: Wait, no, maybe Angle 4 and Angle 5: no, the Vertical Angles would be Angle 1 and Angle 3, Angle 2 and Angle 4, but they are not in the angle pairs. So perhaps the intended matches are:

Angle 1 and Angle 2: Linear Pair

Angle 2 and Angle 5: Complementary Angles

Angle 4 and Angle 5: Complementary Angles (but the descriptions are three, so maybe one of them is Vertical Angles, but the angle pairs don't have vertical angles. Maybe the problem's diagram has Angle 1 and Angle 3 as vertical angles, but the angle pairs given are Angle 1 and Angle 2, Angle 2 and Angle 5, Angle 4 and Angle 5. So perhaps:

Angle 1 and Angle 2: Linear Pair

Angle 2 and Angle 5: Complementary Angles

Angle 4 and Angle 5: Complementary Angles (but the descriptions are Vertical Angles, Linear Pair, Complementary Angles). Maybe the Vertical Angles are not in the given angle pairs, so we proceed with the given.

# Answer:
- Angle 1 and Angle 2: Linear Pair
- Angle 2 and Angle 5: Complementary Angles
- Angle 4 and Angle 5: Complementary Angles (Note: There might be a mis - match in the problem's angle pairs and descriptions, but based on the given, this is the best match.)

### Question 15
# Explanation:
## Step1: Recall the definition of complementary angles
Complementary angles are two angles whose sum is $ 90^{\circ} $. So if $ \angle A $ and $ \angle B $ are complementary, then $ m\angle A + m\angle B=90^{\circ} $.
## Step2: Solve for $ m\angle B $
We know that $ m\angle A = 67^{\circ} $, so we can substitute into the equation: $ 67^{\circ}+m\angle B = 90^{\circ} $. Then, $ m\angle B=90^{\circ}-67^{\circ}=23^{\circ} $.
# Answer:
$ 23^\circ $

### Question 16
# Explanation:
## Step1: Recall the triangle angle - sum property
The sum of the interior angles of a triangle is $ 180^{\circ} $. In $ \triangle NRP $, we know two angles: at $ R $ is $ 34^{\circ} $, at $ Q $ (wait, no, the triangle is $ \triangle PRQ $? Wait, the diagram shows $ \angle NRQ $ is a straight line, so $ \angle NRP $, $ \angle PRQ = 34^{\circ} $, and $ \angle PQR = 95^{\circ} $. Wait, in $ \triangle PRQ $, the sum of angles is $ 180^{\circ} $. So $ m\angle RQP + m\angle QRP+m\angle QPR = 180^{\circ} $. We know $ m\angle RQP = 95^{\circ} $, $ m\angle QRP = 34^{\circ} $, so $ m\angle QPR=180-(95 + 34)=180 - 129 = 51^{\circ} $. Wait, no, the question is $ m\angle NRP $. Wait, $ \angle NRP $ and $ \angle PRQ $ are supplementary? No, $ \angle NRP $ is an exterior angle? Wait, no, $ \angle NRP $ and $ \angle PRQ $ are adjacent angles on a straight line? Wait, no, the diagram: $ N - R - Q $ is a straight line, so $ \angle NRP $ and $ \angle PRQ $ are adjacent, but $ \angle NRP $ is an angle of the triangle? Wait, no, the triangle is $ \triangle PRQ $, with $ \angle at R: 34^{\circ} $, $ \angle at Q:95^{\circ} $. Then, the sum of angles in a triangle is $ 180^{\circ} $, so $ m\angle P=180-(34 + 95)=51^{\circ} $. Wait, no, the question is $ m\angle NRP $. Wait, maybe $ \angle NRP $ is equal to the sum of the two non - adjacent interior angles of the triangle (exterior angle theorem). The exterior angle theorem states that an exterior angle of a triangle is equal to the sum of the two non - adjacent interior angles. So $ \angle NRP $ is an exterior angle at $ R $, so $ m\angle NRP=m\angle P + m\angle Q $. Wait, $ m\angle P $ we found as $ 51^{\circ} $? No, wait, $ m\angle Q = 95^{\circ} $, $ m\angle R = 34^{\circ} $, so $ m\angle P=180 - 95 - 34 = 51^{\circ} $. Then, by exterior angle theorem, $ m\angle NRP=m\angle P + m\angle Q=51 + 95 = 146^{\circ} $? Wait, no, that can't be. Wait, no, the exterior angle at $ R $ would be $ \angle NRP $, and the two non - adjacent interior angles are $ \angle P $ and $ \angle Q $. Wait, no, the exterior angle at $ R $: the interior angles at $ R $ is $ 34^{\circ} $, so the exterior angle $ \angle NRP=180 - 34=146^{\circ} $? Wait, no, $ N - R - Q $ is a straight line, so $ \angle NRP + \angle PRQ=180^{\circ} $? No, $ \angle PRQ = 34^{\circ} $, so $ \angle NRP=180 - 34 = 146^{\circ} $? But that contradicts the exterior angle theorem. Wait, no, I think I made a mistake. Let's start over.

In triangle $ PRQ $:

$ m\angle Q = 95^{\circ} $, $ m\angle R = 34^{\circ} $

Sum of angles in a triangle: $ m\angle P+m\angle Q + m\angle R=180^{\circ} $

So $ m\angle P=180-(95 + 34)=51^{\circ} $

Now, $ \angle NRP $ is an exterior angle to triangle $ PRQ $ at vertex $ R $. The exterior angle theorem states that the measure of an exterior angle of a triangle is equal to the sum of the measures of the two non - adjacent interior angles. So the two non - adjacent interior angles to $ \angle NRP $ are $ \angle P $ and $ \angle Q $.

So $ m\angle NRP=m\angle P + m\angle Q=51^{\circ}+95^{\circ}=146^{\circ} $

Yes, that's correct. Because the exterior angle at $ R $ ( $ \angle NRP $) is equal to the sum of the two remote interior angles ( $ \angle P $ and $ \angle Q $).
## Step2: Calculate using exterior angle theorem
$ m\angle NRP=m\angle P + m\angle Q $

We found $ m\angle P = 51^{\circ} $, $ m\angle Q = 95^{\circ} $, so $ m\angle NRP=51 + 95 = 146^{\circ} $
# Answer:
$ 146^\circ $

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QUESTION IMAGE

Question

Explanation:

Question 13

Step1: Identify the relationship

Step2: Calculate the measure

Step1: Recall the definition of complementary angles

Step2: Solve for \( m\angle B \)

Answer:

Question 14