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# In the figure shown below AB = AC, angle BAD = 30°, and AE = AD. Then

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In the figure shown below AB = AC, angle BAD = 30°, and AE = AD. Then  [#permalink]

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Updated on: 23 Jul 2017, 23:59
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Difficulty:

35% (medium)

Question Stats:

81% (02:39) correct 19% (02:51) wrong based on 56 sessions

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In the figure shown below AB = AC, angle BAD = 30°, and AE = AD. Then x equals:

(A) 7.5°

(B) 10°

(C) 12.5°

(D) 15°

(E) 20°

Attachment:

shot21.jpg [ 19.97 KiB | Viewed 6868 times ]

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Originally posted by carcass on 23 Jul 2017, 23:55.
Last edited by Bunuel on 23 Jul 2017, 23:59, edited 1 time in total.
Edited the question.
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Joined: 30 Mar 2017
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Re: In the figure shown below AB = AC, angle BAD = 30°, and AE = AD. Then  [#permalink]

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24 Jul 2017, 00:02
It's D 15°
Let angle ABD=angleAED=y
So angle DAE = 150-2y
Then angle ADE= Angle AED =15+y

Adding all the angles in triangle DEC ,We get x=15°

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Joined: 02 Sep 2009
Posts: 55230
Re: In the figure shown below AB = AC, angle BAD = 30°, and AE = AD. Then  [#permalink]

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24 Jul 2017, 00:23
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carcass wrote:
In the figure shown below AB = AC, angle BAD = 30°, and AE = AD. Then x equals:

(A) 7.5°

(B) 10°

(C) 12.5°

(D) 15°

(E) 20°

Attachment:
The attachment shot21.jpg is no longer available

AB = AC, means that the red angles are equal.
AE = AD, means that the blue angles are equal.

For triangle ABD: Red + 30° + (180° - Blue - x) = 180°. This gives Red - Blue - x + 30° = 0
For triangle DEC: x + (180° - Blue) + Red = 180°. This gives Red - Blue + x = 0

Equate: Red - Blue - x + 30° = Red - Blue + x
2x = 30.
x = 15°.

Attachment:

Untitled.png [ 16.63 KiB | Viewed 6688 times ]

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Joined: 02 Sep 2009
Posts: 55230
Re: In the figure shown below AB = AC, angle BAD = 30°, and AE = AD. Then  [#permalink]

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24 Jul 2017, 02:37
chigpt wrote:
How ABD = AED??

Sent from my iPhone using GMAT Club Forum

That must be a typo in @BeingHan's solution because Angle ABD = angle ACD. Check here: https://gmatclub.com/forum/in-the-figur ... l#p1893868
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In the figure shown below AB = AC, angle BAD = 30°, and AE = AD. Then  [#permalink]

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Updated on: 24 Jul 2017, 19:46
carcass wrote:
In the figure shown below AB = AC, angle BAD = 30°, and AE = AD. Then x equals:

(A) 7.5°

(B) 10°

(C) 12.5°

(D) 15°

(E) 20°

Attachment:
shot21.jpg

This question is testing exterior angles and related isosceles triangles. For the triangle (ABD) on the left, we can choose an angle for the base of the bigger triangle i.e 50, therefore the external angle would be 80 (30+50) and this would include x. Since we chose 50 for the base angles the third angle of the whole triangle (ABC) would be 180-100= 80, then the third angle of the triangle on the right(ADE) would be 50 (80-30) and the base angles would be 130/2= 65 each. The left base angle of ADE plus x equals 80, then x =80 -65 = 15. Answer is D

Originally posted by rulingbear on 24 Jul 2017, 10:20.
Last edited by rulingbear on 24 Jul 2017, 19:46, edited 1 time in total.
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Joined: 30 May 2013
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GMAT 2: 640 Q49 V29
In the figure shown below AB = AC, angle BAD = 30°, and AE = AD. Then  [#permalink]

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24 Jul 2017, 10:43
1
carcass wrote:
In the figure shown below AB = AC, angle BAD = 30°, and AE = AD. Then x equals:

(A) 7.5°

(B) 10°

(C) 12.5°

(D) 15°

(E) 20°

Attachment:
shot21.jpg

Ans (D) 15
Let angle ABD = Z, since AB=AC, then angle ACB = Z
In Triangle BAC, angle BAC = 180-2z, hence angle DAE = (180-2z) - 30 = 150-2z --------(1)

Now in smaller triangle ADE, let angle ADE = y, since AD=AE, so angle AED = y
y+y+angle DAE = 180
or 2y + (150 -2z) [from (1) ] = 180
or y-z = 15 ------------(2)

Final Step, angle AED (Exterior angle) = Angle EDC + angle ECD (Sum of interior angles)
i.e. y = x + z
or y-z = x = 15 (from (2) )
In the figure shown below AB = AC, angle BAD = 30°, and AE = AD. Then   [#permalink] 24 Jul 2017, 10:43
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