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In the diagram, BC = CD = BD, angle FDE = 40°, angle EFD = 75°, and li

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In the diagram, BC = CD = BD, angle FDE = 40°, angle EFD = 75°, and li  [#permalink]

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New post 18 Jun 2019, 00:49
00:00
A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

80% (02:47) correct 20% (02:30) wrong based on 49 sessions

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Re: In the diagram, BC = CD = BD, angle FDE = 40°, angle EFD = 75°, and li  [#permalink]

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New post 18 Jun 2019, 01:30
Given info:
1) BC = CD = BD
2) angle FED = 40
3) angle EFD = 75
4) BG || DF


Using (1), angles cbd=bcd=bdc=60
Using (2) and (4), and applying concept of Vertically opposite angles:
angle D made by line segment CD is 40.
Angle between extended DF and BD is 60+40 = 100

Applying sum of interior angles on same transverse is 180
Therefore, upper angle B and D sum to 180
angle B + D = 180
=> B= 180-100 = 80

Now, angle B (made by extended line segment BG and CB):
= 80-60 = 20

Again, applying vertically opposite angles:
angle ABG = 20

From (3) and (4);
angle DFE = BGF = 75

angle BGA = 180 - 75 = 105

therefore, angle BAG = 180 - 105 - 20 = 55

Option D as correct answer
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Re: In the diagram, BC = CD = BD, angle FDE = 40°, angle EFD = 75°, and li  [#permalink]

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New post 18 Jun 2019, 01:44
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Bunuel wrote:
Image
In the diagram, BC = CD = BD, angle FDE = 40°, angle EFD = 75°, and line BG is parallel to line DF. What is the measure of angle A in degrees?

A. 25
B. 35
C. 45
D. 55
E. 65


Attachment:
Capture.PNG


Find Angle E using sum of angles in a triangle rule = E=65.
angle C=60 as CBD is equilateral triangle (from stem: BC = CD = BD)
So using sum of angles in a triangle rule ACE arrive at angle A 60+65+A=180; A=55 IMO D
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Re: In the diagram, BC = CD = BD, angle FDE = 40°, angle EFD = 75°, and li  [#permalink]

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New post 18 Jun 2019, 02:19
In this Question, the only property that is to be used is Sum of triangles.

Triangle BCD is an equilateral triangle, hence all the angles are 60. then in triangle, DFE, angle E = 180-75-40, ie., 65
so, triangle AEC, A = 180-65-60
180-125 = 55 ie. D


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Re: In the diagram, BC = CD = BD, angle FDE = 40°, angle EFD = 75°, and li  [#permalink]

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New post 18 Jun 2019, 02:58
Bunuel wrote:
Image
In the diagram, BC = CD = BD, angle FDE = 40°, angle EFD = 75°, and line BG is parallel to line DF. What is the measure of angle A in degrees?

A. 25
B. 35
C. 45
D. 55
E. 65


Attachment:
Capture.PNG


In Triangle DEF, LE +LF + LD = 180
--> LE + 75 + 40 = 180
--> LE = 65

Triangle BCD is Equilateral --> LC = 60

In Triangle ACE, LA + LC + LE = 180
--> LA + 60 + 65 = 180
--> L = 180 - 125 = 55

IMO Option D

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Re: In the diagram, BC = CD = BD, angle FDE = 40°, angle EFD = 75°, and li  [#permalink]

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New post 18 Jun 2019, 10:53
Bunuel wrote:
Image
In the diagram, BC = CD = BD, angle FDE = 40°, angle EFD = 75°, and line BG is parallel to line DF. What is the measure of angle A in degrees?

A. 25
B. 35
C. 45
D. 55
E. 65


Attachment:
Capture.PNG


other side angle is 65 and ∆ BCD is equilateral so 60
180-60+65; 55
angle A is 55
IMO D
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Re: In the diagram, BC = CD = BD, angle FDE = 40°, angle EFD = 75°, and li  [#permalink]

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New post 18 Jun 2019, 12:32
Bunuel wrote:
Image
In the diagram, BC = CD = BD, angle FDE = 40°, angle EFD = 75°, and line BG is parallel to line DF. What is the measure of angle A in degrees?

A. 25
B. 35
C. 45
D. 55
E. 65


Attachment:
Capture.PNG


1. Since in △BCD , BC = CD = BD, its an equilateral Triangle and ∠CBD = ∠BDC = ∠BCD = 60°
2. In △DEF , ∠DEF= 180° - (40° + 75°) => 65°
3. In △ACE, ∠CAE = 180° - (60° + 65°) => 55°

Hence, Answer must be (D)
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Re: In the diagram, BC = CD = BD, angle FDE = 40°, angle EFD = 75°, and li   [#permalink] 18 Jun 2019, 12:32
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