Mathematics
14.08.2021 11:32
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Given that segment ed ∥ segment ab, ∠ced ≅∠cde, m∠cab = 22° , m∠abe = 11°, and m∠acb

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=
Given that segment ed ∥ segment ab, ∠ced ≅∠cde, m∠cab = 22° , m∠abe = 11°, and m∠a
Show Answers
ceceallen2003
ceceallen2003
4,6(85 marks)

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°

cheatcodeb
cheatcodeb
4,9(14 marks)

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°

isabelibarra6370
isabelibarra6370
4,4(100 marks)

Yes, 1x is the same as x. Technically, the "1" in front of the "x" is called an "implied 1" and, so it's simply written as x. Just simply means that any value multiplied by 1 is always equal to itself.

Step-by-step explanation:

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