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D
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Difficulty:
45%
(medium)
Question Stats:
71% (02:43) correct
29%
(02:51)
wrong
based on 63
sessions
History
Date
Time
Result
Not Attempted Yet
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°
shot21.jpg [ 19.97 KiB | Viewed 28295 times ]
(A) 7.5°
(B) 10°
(C) 12.5°
(D) 15°
(E) 20°
Attachment:
shot21.jpg [ 19.97 KiB | Viewed 28295 times ]
24 Jul 2017, 00:23
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carcass
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°
(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°.
Answer: D.
Attachment:
Untitled.png [ 16.63 KiB | Viewed 27472 times ]
General Discussion
24 Jul 2017, 00:02
Kudos
Bookmarks
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°
Sent from my Moto G (5) Plus using GMAT Club Forum mobile app
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°
Sent from my Moto G (5) Plus using GMAT Club Forum mobile app
