## 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.

______Â° + 20Â° = 30Â°

dÂ° = ______Â°

Answer:

The value of dÂ° is 10Â°.

Explanation:

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

dÂ° + 20Â°= 30Â°

so the value of dÂ° is 30Â° – 20Â°

= 10Â°.

Question 2.

______Â° + ______Â°= 360Â°

cÂ° = ______Â°

Answer:

The value of cÂ° is 270Â°.

Explanation:

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

cÂ°= 360Â°-90Â°

= 270Â°.

Question 3.

______Â° + ______Â° + ______Â° = ______Â°

eÂ° = ______Â°

Answer:

The value of eÂ° is 196Â°.

Explanation:

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

eÂ° = 360Â°- 164Â°

= 196Â°.

So the value of eÂ° is 196Â°.

Question 4.

______Â° + ______Â° + ______Â° = ______Â°

fÂ° = ______Â°

Answer:

The value of fÂ° is 110Â°.

Explanation:

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

fÂ° = 360Â°- 250Â°

= 110Â°.

So the value of fÂ° is 110Â°.

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

Question 5.

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

xÂ° = ______Â°Â Â Â yÂ° = ______Â°

Answer:

The value of xÂ° is 160Â°.

The value of yÂ° is 20Â°.

Explanation:

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

yÂ°= 20Â°.

So the value of yÂ° is 20Â°.

And the other equation will be 20Â° + xÂ°= 180Â°

xÂ°= 180Â° – 20Â°

= 160Â°.

So the value of the xÂ°is 160Â°.

Question 6.

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

gÂ° = ______Â°Â Â Â hÂ° = ______Â°Â Â Â Â tÂ° = ______Â°

Answer:

The value of iÂ° is 55Â°.

The value of hÂ° is 125Â°.

The value of gÂ° is 55Â°.

Explanation:

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

iÂ°= 55Â°.

So the value of iÂ° is 55Â°.

And the other equation will be 55Â° + hÂ°= 180Â°

hÂ°= 180Â° – 55Â°

= 125Â°.

So the value of the hÂ°is 125Â°.

And the other equation will be 125Â° + gÂ°= 180Â°

gÂ°= 180Â° – 125Â°

= 55Â°.

So the value of the gÂ°is 55Â°.

Question 7.

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

kÂ° = ______Â°Â Â Â mÂ° = ______Â°Â Â Â nÂ° = ______Â°

Answer:

The value of mÂ° is 54Â°.

The value of kÂ° is 36Â°.

The value of nÂ° is 144Â°.

Explanation:

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

mÂ°= 54Â°.

So the value of mÂ° is 54Â°.

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

kÂ°= 180Â° – 144Â°

= 36Â°.

So the value of the kÂ°is 36Â°.

And the other equation will be 36Â° +nÂ°= 180Â°

nÂ°= 180Â° – 36Â°

= 144Â°.

So the value of the nÂ°is 144Â°.

### 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.

xÂ° =

Answer:

The value of xÂ° is 24Â°.

Explanation:

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

xÂ°= 24Â°.

So the value of xÂ° is 24Â°.

Question 2.

yÂ° =

Answer:

The value of yÂ° is 156Â°.

Explanation:

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

yÂ°= 156Â°.

So the value of yÂ° is 156Â°.

Question 3.

zÂ° =

Answer:

The value of zÂ° is 24Â°.

Explanation:

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

zÂ°= 24Â°.

So the value of zÂ° is 24Â°.

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

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

Question 1.

__________Â° + 320Â° = 360Â°

aÂ° = ________ Â°

Answer:

The value of aÂ° is 40Â°.

Explanation:

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

aÂ° + 320Â°= 360Â°

so the value of aÂ° is 360Â° – 320Â°

= 40Â°.

Question 2.

_____________ Â° + ____________ Â° = 360Â°

bÂ° = __________ Â°

Answer:

The value of bÂ° is 315Â°.

Explanation:

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

bÂ° + 45Â°= 360Â°

so the value of bÂ° is 360Â° – 45Â°

=315Â°

Question 3.

_____________ Â° + _____________ Â° + _____________ Â° = _____________ Â°

cÂ° = _____________ Â°

Answer:

The value of cÂ° is 145Â°.

Explanation:

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

cÂ° + 215Â°= 360Â°

so the value of cÂ° is 360Â° – 215Â°

=145Â°.

Question 4.

_____________ Â° + _____________ Â° + _____________ Â° = _____________ Â°

dÂ° = _____________ Â°

Answer:

The value of dÂ° is 80Â°.

Explanation:

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

dÂ° + 280Â°= 360Â°

so the value of cÂ° is 360Â° – 280Â°

=80Â°.

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

Question 5.

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

eÂ° = _____________ Â°Â Â Â Â fÂ° = _____________ Â°

Answer:

The value of eÂ° is 145Â°.

The value of yÂ° is 35Â°.

Explanation:

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

eÂ°= 145Â°.

So the value of eÂ° is 145Â°.

And the other equation will be 145Â° + fÂ°= 180Â°

xÂ°= 180Â° – 145Â°

= 35Â°.

Question 6.

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

gÂ° = _____________ Â°Â Â Â Â Â hÂ° = _____________ Â°Â Â Â Â iÂ° = _____________ Â°

Answer:

The value of gÂ° is 125Â°.

The value of hÂ° is 125Â°.

The value of iÂ° is 55Â°.

Explanation:

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

gÂ°= 125Â°.

So the value of gÂ° is 125Â°.

And the other equation will be 55Â° + hÂ°= 180Â°

hÂ°= 180Â° – 55Â°

= 125Â°.

So the value of the hÂ°is 125Â°.

And the other equation will be 125Â° +gÂ°= 180Â°

gÂ°= 180Â° – 125Â°

= 55Â°.

So the value of the gÂ°is 55Â°.

Question 7.

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

JÂ° = _____________ Â°Â Â Â Â Â KÂ° = _____________ Â°Â Â Â Â Â mÂ° = _____________ Â°

Answer:

The value of jÂ° is 44Â°.

The value of kÂ° is 46Â°.

The value of mÂ° is 134Â°.

Explanation:

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

jÂ°= 44Â°.

So the value of jÂ° is 44Â°.

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

kÂ°= 180Â° – 134Â°

= 46Â°.

So the value of the kÂ°is 46Â°.

And the other equation will be 46Â° +mÂ°= 180Â°

mÂ°= 180Â° – 46Â°

= 134Â°.

So the value of the mÂ°is 134Â°.