Practice the problems of Math in Focus Grade 7 Workbook Answer Key Chapter 6 Lesson 6.2 Angles that Share a Vertex to score better marks in the exam.
Math in Focus Grade 7 Course 2 B Chapter 6 Lesson 6.2 Answer Key Angles that Share a Vertex
Math in Focus Grade 7 Chapter 6 Lesson 6.2 Guided Practice Answer Key
Complete.
Question 1.
Find the value of p in the diagram.
Answer:
p° = 20°
Explanation:
7 p°+ 75° + 145° = 360°
7 p° + 220° = 360°
7p° + 220° – 360° = 360°-360°
7p° = 140°
p° = 140 ÷ 7
p° = 20°
Solve.
Question 2.
\(\overleftrightarrow{B E}\) and \(\overleftrightarrow{C A}\) are straight lines. Find the value of each variable.
Answer:
r°=28°, q°=28°, 3q°=84°
Explanation:
∠BOA + ∠AOE = 180°
152°+r°=180°
r = 180°-152°
r= 28°
∠BOC + ∠COD + ∠DOE = 180°
q°+ 68°+3q° = 180°
4q°+68° = 180°
4q° = 180°-68°
4q° = 112°
q°= 112÷4
q°= 28°
3q°= 84°
Question 3.
\(\overleftrightarrow{P Q}\) is a straight line. Find the value of each variable.
Answer:
m°=31°, n°=17°, 3m°=93°
Explanation:
m°+149° = 180°
m° = 180°-149°
m° = 31°
3m°= 93°
n°+3m°+70°=180°
n°+93°+70°=180°
n°=180°-163°
n°= 17°
Complete.
Question 4.
In the diagram at the right, the ratio a: b: c = 1: 3 : 5. Find the values of a, b, and c.
The ratio a : b : c = 1 : 3 : 5. So, b = 3 • a and c = 5 • a.
Answer:
a°= 40°, b°=120°, c°=200°
Explanation:
The ratio given is a : b : c = 1 : 3 : 5. Let a=1a, b=3a, c=5a
Technology Activity
Materials:
- geometry software
Explore The Relationship Among Vertical Angles Using Geometry Software
Work in pairs.
Step 1.
Construct intersecting line segments, \(\overline{A B}\) and \(\overline{C D}\), as shown.
Step 2.
Select ∠AOD and find its measure.
Step 3.
Select ∠COB and find its measure.
Step 4.
Then select ∠AOC and ∠BOD and find their measures.
Step 5.
Select point D and drag it so that you change the measures of ∠AOD and ∠AOC. As the measure of ∠AOD changes, what do you notice about the measure of ∠AOC?
Math Journal
Describe what you notice about the measures of the vertical angles.
Answer:
Vertical angles are always equal to one another and are always congruent. The four angles all together always sum to a full angle 360°.
Complete.
Question 5.
\(\overleftrightarrow{A B}\) and \(\overleftrightarrow{C D}\) are straight lines. Find the value of y.
Answer:
3y°=120°
Explanation:
∠COB = ∠AOD
3y°=120°
y°= 120°÷3
y° = 40°
Complete.
Question 6.
In the diagram, two straight lines intersect to form angles 1, 2, 3, and 4. Find the value of each variable if m∠1 = 114°.
m∠1 + m∠2 = 180° Supp. ∠S
Answer:
r°=66°, p°=66°, q°=114°
Explanation:
114°+r°=180°
r°=180°-114
r°=66°
∠3=∠1
q°=114°
∠4=∠2
p°=r°
as r°=66°
p°=66°
Math in Focus Course 2B Practice 6.2 Answer Key
Find the value of each variable.
Question 1.
Answer:
x°=307°
Explanation:
x°+53°=360°
x°=360°-53°
x°=307°
Question 2.
Answer:
x°=257°
Explanation:
x°+33°=290°
x°=290°-33°
x°=257°
Question 3.
Answer:
x°=100°
Explanation:
x°+110°+150°=360°
x°+260°=360°
x°=100°
Question 4.
Answer:
x°=95°
Explanation:
∠AOD+∠BOC+∠AOD = 360°
x°+110°+95°+60°=360°
x°+265°=360°
x°=360°-265°
x°=95°
Question 5.
\(\overleftrightarrow{A B}\) and \(\overleftrightarrow{C D}\) are straight lines.
Answer:
p°= 34°
Explanation:
∠COA = ∠BOD
34° = p°
Hence p° = 34°
Question 6.
\(\overleftrightarrow{A B}\) and \(\overleftrightarrow{C D}\) are straight lines.
Answer:
r°=16°
Explanation:
In the image given below, we can observe that CD and AB are two straight lines. Here, ∠AOB and ∠COD are vertical angles.
a°= 164°
a°+r°=180°
164°+r°=180°
r°=180°-164°
r°=16°
Question 7.
\(\overleftrightarrow{A B}\), \(\overleftrightarrow{C D}\) and \(\overleftrightarrow{E F}\) are straight lines.
Answer:
a° = 35°, b° = 106°
Explanation:
∠FOD = ∠COE
a° = 35° (vertically opposite angles)
∠EOB = ∠AOF
b° = 106° (Vertically opposite angles)
Question 8.
\(\overleftrightarrow{A B}\), \(\overleftrightarrow{C D}\) and \(\overleftrightarrow{E F}\) are straight lines.
Answer:
b° = 126°, a° = 31°
Explanation:
∠AOF = ∠EOB
b° = 126° (vertically opposite angles)
∠AOC = ∠DOB
31° = a° (vertically opposite angles)
Question 9.
\(\overleftrightarrow{A B}\), \(\overleftrightarrow{C D}\) and \(\overleftrightarrow{E F}\) are straight lines.
Answer:
x° = 119°, y°= 122°
Explanation:
∠COE + ∠AOC = 180°
y°+ 58° = 180°
y° = 180°-58°
y° = 122°
∠DOB + ∠FOD = 180°
x° + 61° = 180°
x° = 180°-61°
x° = 119°
Question 10.
\(\overleftrightarrow{A B}\) and \(\overleftrightarrow{C D}\) are straight lines.
Answer:
w + 90° = 168° Vertical ∠s:
w° + 90° – 90° = 168° – 90° Subtract 90° from both sides:
w° = 78° Simplify:
w = 78
Find the value of k.
Question 11.
The ratio ∠1 : ∠2 : ∠3 : ∠4 = 3 : 2 : 1 : 3.
Answer:
k° = 40°
Explanation:
Let the assume given ratio be x
Now, the ratio is 3x: 2x: x: 3x
The sum of all angles at one point is 360°
Now add 3x+ 2x + x + 3x = 360°
9x = 360°
x = 360 ÷ 9
x° = 40°
3x° = 120°
2x° = 80°
3x° = 120°
Now we need to calculate the numbered angle 3 which is named as k° that is 40°
Question 12.
\(\overleftrightarrow{A B}\) and \(\overleftrightarrow{C D}\) are straight lines.
Answer:
4k°= 84°, 3k° = 63
Explanation:
∠COE +∠EOF = 174°
4k°+ 90° = 174°
4k° = 174-90
4k°= 84°
k° = 84÷4
k° = 21°
3k°= 21×3
3k° = 63°
Question 13.
Answer:
2k°+5k°+3k°+130°=180°
10k°+130°=360°
10k° = 360°-130°
10k° = 230°
k° = 23°
IF k°=23° then 2k° = 2×23 = 46°, 3k° = 3×23=69°
5k° = 5×23=115°
Name the pairs of vertical angles.
Question 14.
Answer:
AE and DB are a pair of non-adjacent angles that are formed when two lines intersect.
Explanation:
Vertical lines are a pair of non-adjacent angles formed when two lines intersect.
Question 15.
Answer:
Vertical angles are a pair of opposite angles formed by intersecting lines.
∠MKF and ∠NKG Lines \(\overline{M N}\) and \(\overline{F G}\) intersect in K.
∠FKN and ∠MKC We identify the pairs of vertical angles:
∠MKF and ∠NKG
∠FKN and ∠MKC
Question 16.
\(\overleftrightarrow{P S}\) and \(\overleftrightarrow{R N}\) are straight lines.
Answer:
Vertical angles are a pair of opposite angles formed by ¡ntersecting tines.
∠SOR ard ∠NOP Lines \(\overline{P S}\) and \(\overline{R N}\) intersect in O.
∠SON and ∠ROP We identify the pairs of vertical angles:
∠SOR and ∠NOP
∠SON ard ∠ROP
Find the value of each variable.
Question 17.
Answer:
64°+23°+3e°=360°
87°+3e°=360°
3e°=360°-87°
3e°=273°
e°=273°÷ 3°
e° = 91°
Question 18.
\(\overleftrightarrow{A B}\) and \(\overleftrightarrow{C D}\) are straight lines.
Answer:
2p°=88, q°= 92°, r°=2°
Explanation:
∠DOA = ∠BOC
2p°=88
p°= 88÷2
p°= 44
∠DOE+∠EOB+∠BOC= 180°
r°+2p°+88 = 180°
r°+88+88 = 180°
r°+178°=180°
r°=180°-178°
r°=2°
2p°+q°=180°
88+q°=180°
q°=180-88
q°= 92°
Question 19.
\(\overleftrightarrow{A B}\), \(\overleftrightarrow{C D}\) and \(\overleftrightarrow{E F}\) are straight lines. The ratio a : b : c = 1 : 2 : 2.
Answer:
a° = 36°, b = 72°, c = 72°
Explanation:
Let the ratio be x
The sum of the ratio be x + 2x + 2x = 5x
∠AOC = ∠DOB (vertically opposite angles are equal)
∠FOD + ∠DOB + ∠EOB = 180°
c°+a°+b° = 180°
2x + x° + 2x° = 180°
5x° = 180°
x° = 180÷5
x° = 36°
2x°= 72°
Question 20.
\(\overleftrightarrow{A B}\) and \(\overleftrightarrow{C D}\) are straight lines.
Answer:
f° = 40° Vertical ∠s:
f = 40
e° + f° = 180° Straight ∠:
e° + 40° = 1800 Substitute:
e° + 40° – 40° = 180° – 40° Subtract 40° from both sides:
e = 140 Simplify:
d° + 90° + 15° = e° Vertical ∠s:
d° + 105° = 140° Simplify:
d° + 105° — 105° = 140° – 105° Subtract 105° from both sides:
d = 35 Simplify:
f = 40
e = 140
d = 35
Answer each of the following.
Question 21.
In the diagram, the ratio p: q: r = 1: 2 : 3. Find the values of p, q, and r.
Answer:
p = 60°, q = 120°, r = 180°
Explanation:
p: q: r = 1: 2 : 3
x+2x+3x = 360°
6x = 360°
x = 360° ÷ 6
x = 60°
2x= 120°
3x = 180°
p = 60°, q = 120°, r = 180°
Question 22.
If ∠P and ∠N are angles at a point and m∠P = 149°, what is the m∠N?
Answer:
m∠N = 31°
Explanation:
The sum of angles at a point is 180°
Given angles are ∠P = 149°
Here we need to calculate ∠N
∠P + ∠N = 180°
149° + ∠N = 180°
∠N = 180°-149°
∠N = 31°
Question 23.
If 67°, 102°, 1 5°, and x° are angles at a point, what is the value of x?
Answer:
x° = 176°
Explanation:
67°+102°+15°+x°=360°
184°+x° = 360°
x°=360° -184°
x° = 176°
In the diagram below, \(\overleftrightarrow{M P}\) and \(\overleftrightarrow{Q R}\) are straight lines. Answer each of the following.
Question 24.
Name the angle that is vertical to ∠MNR.
Answer:
∠QNP
Explanation:
Vertical angles mean the angles that are opposite to each other. By observing from the given figure the angle ∠QNP is vertical to ∠MNR.
Question 25.
What kind of angles are ∠RNP and ∠PNS?
Answer:
∠RNP and ∠PNS are adjacent angles because they have a common vertex (N) and a common side (\(\overline{N P}\)) and they do not overlap.
Adjacent angles
Question 26.
Find the measure of ∠QNS.
Answer:
m∠QNS + m∠SNP = m∠MNR Vertical ∠s:
m∠QNS + 61° = Substitute:
m∠QNS + 61° – 61° = 138° – 61° Subtract 61° from both sides:
m∠QNS = 77° Simplify:
m∠QNS = 77°
Question 27.
Find the measure of ∠PNR.
Answer:
m∠MNP = 180° Straight ∠
m∠PNR + 138° = 180° Substitute:
m∠PNR + 138° — 138° = 180° — 138° Subtract 138° from both sides:
m∠PNR = 42° Simplify:
m∠PNR = 42°
Use an equation to find the value of each variable.
Question 28.
\(\overleftrightarrow{\mathrm{AB}}\) is a straight line.
Answer:
x°=30°, y°=150°, w°=120°
Explanation:
∠COE + ∠BOE =90°
2x°+x°= 90°
3x°=90
x°=30°
2x°=60°
∠AOC + ∠COE = 180°
w°+2x°=180°
w°+60°=180°
w°=180°-60°
w°=120°
∠AOE + ∠BOE = 180°
y°+x°=180°
y°+30°=180°
y°=180°-30°
y°=150°
Question 29.
\(\overleftrightarrow{\mathrm{AB}}\) and \(\overleftrightarrow{\mathrm{CD}}\) is a straight line.
Answer:
3b° = 90°, b° = 30°, c° = 60°
Explanation:
3b°=90°
b°=90÷3
b°=30°
∠DOB = ∠AOC (vertically opposite angles)
∠AOC = c
∠DOB = c°
∠AOD = 120°
∠DOB + ∠AOD = 180°
c°+120° = 180°
c° = 180°- 120°
c° = 60°
\(\overleftrightarrow{\mathrm{AB}}\) and \(\overleftrightarrow{\mathrm{CD}}\) are straight lines.
Question 30.
The ratio ∠1 : ∠2 = 3 : 1
Answer:
p°= 45°, q°= 135°
Explanation:
Let the given ratio be x
The sum of angles in a traingle is 180°
∠AOD +∠AOC = 180°
The ratio ∠1 : ∠2 = 3x : 1x
3x + x = 180°
4x = 180°
x°= 180÷4
x°= 45°
3x = 135°
∠1 = ∠3 vertically opposite angles
∠2 = ∠4 vertically opposite angles
∠3 = q°= 135°
∠4= p°= 45°
Question 31.
Answer:
p°=50°, q°= 80°, 2p° = 100°
Explanation:
∠AOC = ∠DOB
2p° = p+50
2p°-p° = 50
p°=50°
∠DOB = 100°
∠DOB+∠COB = 180°
P+50 = 50+50 = 100°
100°+q°=180°
q°=180°-100°
q°= 80°
Solve.
Question 32.
The diagram below shows the flag of the Philippines, m∠ADB = 60° and m∠ADC = m∠BDC. Find the measures of ∠ADC and ∠BDC.
Answer:
∠ADB = 60°, ∠BDC = 150°
Explanation:
∠ADB = 60°
∠ADC = ∠BDC
∠ADB + ∠ADC +∠BDC = 360°
60° + ∠ADC + ∠ADC = 360°
60 + 2 ∠ADC = 360°
2 ∠ADC = 360°-60°
2∠ADC = 300°
∠ADC = 300÷2
∠ADC = 150°
∠BDC = 150°
Question 33.
Math Journal
The diagram shows a pattern on a carpet.
a) Are ∠4 and ∠6 vertical angles? Explain why or why not.
Answer:
No.
When two lines cross then vertical angles are opposite to each other.
Explanation:
The angles ∠4 and ∠6 are not vertical angles. Because the vertical angles are opposite to each other but by observing the angles from the given figure the angles given are not opposite to each other.
b) Suppose m∠4 = m∠6. Are ∠4 and ∠5 supplementary angles? Explain your answer.
Answer:
Yes, angles ∠4 and ∠5 are supplementary angles.
Explanation:
Two angles are supplementary when if they add up to 180°. If we add up the given angles ∠4 and ∠5 then the angles add up to 180°