Given that segment ed ∥ segment ab, ∠ced
≅∠cde, m∠cab = 22°
, m∠abe = 11°, and
m∠acb = 136°: fill in all of the missing angles
on this image.
m∠ecd=
m∠eba=
m∠cde=
m∠eab=
m∠ced=
m∠aeb=
m∠deb=
m∠cba=
m∠edb=
m∠dbe=

With the angles given in the triangle, the missing angles on the image are:

1.) m∠ECD = 136°  

2.) m∠EBA = 11°

3.) m∠CDE =  22°

4.) m∠EAB =  22°

5.) m∠CED =  22°

6.) m∠AEB =  147°

7.) m∠DEB =  11°

8.) m∠CBA =  22°

9.) m∠EDB =  158°

10.) m∠DBE = 11°

Step-by-step explanation:

Segment ED ∥ segment AB

∠CED ≅ ∠CDE

m∠CAB = 22°

m∠ABE = 11°

m∠ACB = 136°

1.) m∠ECD

The m∠ECD is the same that the  m∠ACB, then:

m∠ECD = m∠ACB

m∠ECD = 136°

2.) m∠EBA

The m∠EBA is the same that the  m∠ABE, then:

m∠EBA = m∠ABE

m∠EBA = 11°

3.) m∠CDE

It is given that:

∠CED ≅ ∠CDE, then  m∠CED = m∠CDE

And because of the segment ED ∥ segment AB, the ∠CED must be congruent with  ∠CAB, because the corresponding angles in parallel lines (ED and AB) cut by a secant (CA) must be congruent, then:

m∠CED = m∠CAB

m∠CED = 22° (Answer part 5)

m∠CED = m∠CDE

22° = m∠CDE

m∠CDE = 22°

4.) m∠EAB

The m∠EAB is the same that the  m∠CAB, then:

m∠EAB = m∠CAB

m∠EAB = 22°

5.) m∠CED

See Explanation in part 3:

m∠CED = 22°

6.) m∠AEB

In triangle AEB:

m∠EAB = 22°

m∠ABE = 11°

m∠AEB = ?

The sum of the measurements of the interior angles of any triangle must be equal to 180°, then:

m∠EAB + m∠ABE + m∠AEB = 180°

Replacing the known values in the equation above:

22° + 11° +  m∠AEB = 180°

Adding like terms on the left side of the equation:

33° + m∠AEB = 180°

Solving for m∠AEB: Subtracting 33° both sides of the equation:

33° + m∠AEB - 33° = 180° - 33°

Subtracting:

m∠AEB = 147°

7.) m∠DEB

Because of the segment ED // segment AB, the <DEB must be congruent with <ABE, because the alternate interior angles in parallel lines (ED and AB) cut by a secant (BE) must be congruent, then:

m<DEB = m<ABE

m<DEB = 11°

8.) m∠CBA

Because of the segment ED // segment AB, the <CBA must be congruent with <CDE, because the corresponding angles in parallel lines (ED and AB) cut by a secant (BC) must be congruent, then:

m<CBA = m<CDE (=22°, see part 3)

m<CBA = 22°

9.) m∠EDB

According with the figure, and because BC is a straight line, the sum of the measurements of the <CDE and <EDB must be equal to 180°, then:

m<CDE + m<EDB = 180°

By part 3 we know that m<CDE=22°. Replacing this value in the equation above:

22° + m<EDB = 180°

Solving for m<EDB: Subtracting 22° both sides of the equation:

22° + m<EDB - 22° = 180° - 22°

m<EDB = 158°

10.) m∠DBE

In triangle DBE:

m∠EDB = 158° (see part 9)

m∠DEB = 11° (see part 7)

m∠DEB = ?

The sum of the measurements of the interior angles of any triangle must be equal to 180°, then:

m∠EDB + m∠DEB + m∠DEB = 180°

Replacing the known values in the equation above:

158° + 11° +  m∠DEB = 180°

Adding like terms on the left side of the equation:

169° + m∠DEB = 180°

Solving for m∠DEB: Subtracting 169° both sides of the equation:

169° + m∠DEB - 169° = 180° - 169°

Subtracting:

m∠DEB = 11°

m∠ECD = 136°

m∠EBA = 11°

m∠CDE = 22°

m∠EAB = 22°

m∠CED = 22°

m∠AEB = 147°

m∠DEB = 11°

m∠CBA = 22°

m∠EDB = 158°

m∠DBE = 11°

Step-by-step explanation:

Clearly, m∠ECD = m∠ACB

Therefore, m∠ECD = 136°

Clearly, m∠EBA = m∠ABE

Therefore, m∠EBA = 11°

m∠CDE ≅ m∠CED (given)

But, since segment ED || segment AB,

m∠CED ≅ m∠CAB

But, m∠CAB =  22° (given)

So,  m∠CDE =  22°

Note that m∠EAB = m∠CAB

Therefore, m∠EAB = 22°

m∠CED = m∠CDE = 22°

In ΔEAB,

m∠EAB + m∠EBA + m∠AEB = 180°

22 + 11 + m∠AEB = 180

33 + m∠AEB = 180

m∠AEB = 180 - 33

m∠AEB = 147°  

Note that m∠DEB = m∠EBA (Alternate interior angles)

Therefore, m∠DEB = 11°

m∠CAB = m∠CED and

m∠CBA = m∠CDE (Corresponding angles)

But, it is given that m∠CED ≅ m∠CDE

Therefore, m∠CAB = m∠CED = m∠CDE = m∠CBA

and hence m∠CBA = 22°

In ΔEDB,

m∠EDB + m∠DBE + m∠BED = 180° (Angle sum property)

m∠EDB + 11 + 11 = 180

m∠EDB = 158°

We know that m∠CBA = 22° and m∠ABE = 11°

m∠DBE = m∠CBA - m∠ABE

= 22 - 11

11

Hence, m∠DBE = 11°

the distance between the school and her home is 3.6km.

12 mins = 0.2 hours = 1/5 hour

54 ÷ 3 = 18km per hour

1/5 of 18 = 3.6km per 12 mins