## Engage NY Eureka Math 4th Grade Module 4 Lesson 11 Answer Key

### Eureka Math Grade 4 Module 4 Lesson 11 Problem Set Answer Key

Write an equation, and solve for the unknown angle measurements numerically.

Question 1.

______° + 20° = 30°

d° = ______°

Answer:

The value of d° is 10°.

Explanation:

Given that the value of the angle acute angle is 30° and the value of the other angle is 20° and the value of another angle is d°. So the equation will be

d° + 20°= 30°

so the value of d° is 30° – 20°

= 10°.

Question 2.

______° + ______°= 360°

c° = ______°

Answer:

The value of c° is 270°.

Explanation:

Here, in the above image, we can see that an arc that represents a complete rotation which means 360°, and the other angle 90°. So the equation will be c°+90°= 360° and the value of c° is

c°= 360°-90°

= 270°.

Question 3.

______° + ______° + ______° = ______°

e° = ______°

Answer:

The value of e° is 196°.

Explanation:

Here we will measure the angles using a protractor and the values of the angles will be 90° and 196°. So the equation will be 74°+90°+e°= 360° and the value of e is

e° = 360°- 164°

= 196°.

So the value of e° is 196°.

Question 4.

______° + ______° + ______° = ______°

f° = ______°

Answer:

The value of f° is 110°.

Explanation:

Here we will measure the angles using a protractor and the values of the angles will be 90° and 160°. So the equation will be 90°+160°+f°= 360° and the value of f is

f° = 360°- 250°

= 110°.

So the value of f° is 110°.

Write an equation, and solve for the unknown angles numerically.

Question 5.

O is the intersection of \(\overline{A B}\) and \(\overline{C D}\). ∠DOA is 160°, and ∠AOC is 20°.

x° = ______° y° = ______°

Answer:

The value of x° is 160°.

The value of y° is 20°.

Explanation:

In the above image, we can see that the angle COD is 180° and <DOA is 160° and <AOC is 20°. So the equation will be 160° + y°= 180° and the value of y°= 180° – 160°

y°= 20°.

So the value of y° is 20°.

And the other equation will be 20° + x°= 180°

x°= 180° – 20°

= 160°.

So the value of the x°is 160°.

Question 6.

O is the intersection of \(\overline{R S}\) and \(\overline{T V}\). ∠TOS is 125°.

g° = ______° h° = ______° t° = ______°

Answer:

The value of i° is 55°.

The value of h° is 125°.

The value of g° is 55°.

Explanation:

In the above image, we can see that the angle TOV is 180° and angle SOR is 180° and <TOS is 125°. So the equation will be 125° + i°= 180° and the value of i°= 180° – 125°

i°= 55°.

So the value of i° is 55°.

And the other equation will be 55° + h°= 180°

h°= 180° – 55°

= 125°.

So the value of the h°is 125°.

And the other equation will be 125° + g°= 180°

g°= 180° – 125°

= 55°.

So the value of the g°is 55°.

Question 7.

O is the intersection of \(\overline{W X}\), \(\overline{Y Z}\), and \(\overline{U O}\) ∠XOZ is 36°

k° = ______° m° = ______° n° = ______°

Answer:

The value of m° is 54°.

The value of k° is 36°.

The value of n° is 144°.

Explanation:

In the above image, we can see that the angle WOX is 180° and angle XOZ is 36°, and the value of angle UOX is 90°. So the equation will be 90°+m+ 36°= 180° and the value of m°= 180° – 126°

m°= 54°.

So the value of m° is 54°.

And the other equation will be 54° +90°+k°= 180°

k°= 180° – 144°

= 36°.

So the value of the k°is 36°.

And the other equation will be 36° +n°= 180°

n°= 180° – 36°

= 144°.

So the value of the n°is 144°.

### Eureka Math Grade 4 Module 4 Lesson 11 Exit Ticket Answer Key

Write equations using variables to represent the unknown angle measurements. Find the unknown angle measurements numerically.

Question 1.

x° =

Answer:

The value of x° is 24°.

Explanation:

In the above image, we can see that the angle AEB is 180° and angle AEF is 90°. So the equation will be 90°+x+66= 180° and the value of m°= 180° – 156°

x°= 24°.

So the value of x° is 24°.

Question 2.

y° =

Answer:

The value of y° is 156°.

Explanation:

In the above image, we can see that the angle AEB is 180° and angle AEF is 90°. So the equation will be 24°+y= 180° and the value of y°= 180° – 24°

y°= 156°.

So the value of y° is 156°.

Question 3.

z° =

Answer:

The value of z° is 24°.

Explanation:

In the above image, we can see that the angle AEB is 180° and angle FEB is 90°. So the equation will be 90°+z+66= 180° and the value of m°= 180° – 156°

z°= 24°.

So the value of z° is 24°.

### Eureka Math Grade 4 Module 4 Lesson 11 Homework Answer Key

Write an equation, and solve for the unknown angle measurements numerically.

Question 1.

__________° + 320° = 360°

a° = ________ °

Answer:

The value of a° is 40°.

Explanation:

Given that the value of the angle is complete rotation which 360° and the value of the other angle is 320° and the value of another angle is a°. So the equation will be

a° + 320°= 360°

so the value of a° is 360° – 320°

= 40°.

Question 2.

_____________ ° + ____________ ° = 360°

b° = __________ °

Answer:

The value of b° is 315°.

Explanation:

Given that the value of the angle is complete rotation which 360° and the value of the other angle is 45° and the value of another angle is b°. So the equation will be

b° + 45°= 360°

so the value of b° is 360° – 45°

=315°

Question 3.

_____________ ° + _____________ ° + _____________ ° = _____________ °

c° = _____________ °

Answer:

The value of c° is 145°.

Explanation:

Given that the value of the angle is complete rotation which 360° and the value of the other angle is 115° and the value of another angle is 100° and the value of another angle is c°. So the equation will be c°+115°+100°= 360°.

c° + 215°= 360°

so the value of c° is 360° – 215°

=145°.

Question 4.

_____________ ° + _____________ ° + _____________ ° = _____________ °

d° = _____________ °

Answer:

The value of d° is 80°.

Explanation:

Given that the value of the angle is complete rotation which 360° and the value of the other angle is 135° and the value of another angle is 145° and the value of another angle is d°. So the equation will be d°+135°+145°= 360°.

d° + 280°= 360°

so the value of c° is 360° – 280°

=80°.

Write an equation, and solve for the unknown angles numerically.

Question 5.

O is the intersection of \(\overline{A B}\) and \(\overline{C D}\). ∠COB is 145°, and ∠AOC is 35°

e° = _____________ ° f° = _____________ °

Answer:

The value of e° is 145°.

The value of y° is 35°.

Explanation:

In the above image, we can see that the angle COD is 180° and <COB is 145° and <AOC is 35°. So the equation will be 35° + e°= 180° and the value of e°= 180° – 35°

e°= 145°.

So the value of e° is 145°.

And the other equation will be 145° + f°= 180°

x°= 180° – 145°

= 35°.

Question 6.

O is the intersection of \(\overline{Q R}\) and \(\overline{S T}\). ∠QOS is 55°

g° = _____________ ° h° = _____________ ° i° = _____________ °

Answer:

The value of g° is 125°.

The value of h° is 125°.

The value of i° is 55°.

Explanation:

In the above image, we can see that the angle SOT is 180° and angle QOR is 180° and <QOS is 55°. So the equation will be 55° + g°= 180° and the value of g°= 180° – 55°

g°= 125°.

So the value of g° is 125°.

And the other equation will be 55° + h°= 180°

h°= 180° – 55°

= 125°.

So the value of the h°is 125°.

And the other equation will be 125° +g°= 180°

g°= 180° – 125°

= 55°.

So the value of the g°is 55°.

Question 7.

O is the intersection of \(\overline{U V}\), \(\overline{W X}\), and \(\overline{Y O}\). ∠VOX is 46°

J° = _____________ ° K° = _____________ ° m° = _____________ °

Answer:

The value of j° is 44°.

The value of k° is 46°.

The value of m° is 134°.

Explanation:

In the above image, we can see that the angle WOX is 180° and angle UOV is 180° and <VOX is 46°. So the equation will be 46° +90°+ j°= 180° and the value of j°= 180° – 136°

j°= 44°.

So the value of j° is 44°.

And the other equation will be 44° +90°+ k°= 180°

k°= 180° – 134°

= 46°.

So the value of the k°is 46°.

And the other equation will be 46° +m°= 180°

m°= 180° – 46°

= 134°.

So the value of the m°is 134°.