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In the figure below, ∠ADE = 60°, ∠EFC = 40°, and ∠DAE = 55°. If AB ||

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In the figure below, ∠ADE = 60°, ∠EFC = 40°, and ∠DAE = 55°. If AB ||  [#permalink]

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New post 20 Mar 2015, 07:13
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Re: In the figure below, ∠ADE = 60°, ∠EFC = 40°, and ∠DAE = 55°. If AB ||  [#permalink]

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New post 20 Mar 2015, 12:31
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its A =75

angle AED=65 (AED is 180) =angle FEC( opp)

thru above FCE =75 which makes ECB=115 again

EAB =65(vertically opp angles thru thr trnsversal between the parallel lines)
now we have all four angles of quad ABCE

so we can solve x=75
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Re: In the figure below, ∠ADE = 60°, ∠EFC = 40°, and ∠DAE = 55°. If AB ||  [#permalink]

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New post 20 Mar 2015, 20:35
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Bunuel wrote:
Attachment:
originals (1).jpg
In the figure below, ∠ADE = 60°, ∠EFC = 40°, and ∠DAE = 55°. If AB || CD, what is the value of x? Note: Figure not drawn to scale

A. 75
B. 85
C. 95
D. 105
E. 115

Kudos for a correct solution.


It looks to me that triangles ABF and ECF are similar triangles.
If that is the case then angle AED is 65 (found by 180-55-60=65)
Angle CEF is a vertical angle so it also equals 65.
Finally this leads us to angle ECF which is 75 (found by 180-65-40=75)

x= 75

Answer: A
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figure below, ∠ADE = 60°, ∠EFC = 40°, and ∠DAE = 55°. If AB ||  [#permalink]

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New post 23 Mar 2015, 00:29
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Answer = A = 75

Refer diagram below:

Attachment:
originals%20(1).jpg
originals%20(1).jpg [ 20.85 KiB | Viewed 3627 times ]


\(\angle ABC = \angle ECF = x\) (They are alternate angles)

x = 180-(75+40) = 75
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Re: In the figure below, ∠ADE = 60°, ∠EFC = 40°, and ∠DAE = 55°. If AB ||  [#permalink]

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New post 23 Mar 2015, 05:43
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Re: In the figure below, ∠ADE = 60°, ∠EFC = 40°, and ∠DAE = 55°. If AB ||  [#permalink]

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New post 23 Mar 2015, 11:14
Bunuel wrote:
Attachment:
originals (1).jpg
In the figure below, ∠ADE = 60°, ∠EFC = 40°, and ∠DAE = 55°. If AB || CD, what is the value of x? Note: Figure not drawn to scale

A. 75
B. 85
C. 95
D. 105
E. 115

Kudos for a correct solution.



Answer is A: 75.

Number of properties come into play.
First, sum of angles in a triangle is 180. That gives us value of \(<AED = 65.\)
Second, Vertically opposite angles are equal, gives us \(<AED = <CEF = 65\).
Again, angle sum property of a Triangle gives \(<ECF = 75\)

Third, Linear pair, gives \(< BCD = 105\)

Fourth, Sum of angles on the same side of transversal BC (AB parallel CD) is supplementary, gives \(<ABC=x=75\).

Thank you.
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Re: In the figure below, ∠ADE = 60°, ∠EFC = 40°, and ∠DAE = 55°. If AB ||  [#permalink]

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New post 25 Mar 2016, 08:37
Bunuel wrote:
Attachment:
originals (1).jpg
In the figure below, ∠ADE = 60°, ∠EFC = 40°, and ∠DAE = 55°. If AB || CD, what is the value of x? Note: Figure not drawn to scale

A. 75
B. 85
C. 95
D. 105
E. 115

Kudos for a correct solution.


AED=CEF=65 deg

ECF=75 deg

ECF=ABC (Corresponding angles)
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Re: In the figure below, ∠ADE = 60°, ∠EFC = 40°, and ∠DAE = 55°. If AB ||  [#permalink]

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Re: In the figure below, ∠ADE = 60°, ∠EFC = 40°, and ∠DAE = 55°. If AB ||   [#permalink] 18 Jul 2019, 10:40
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