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04 Jun 2013, 11:27
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
85% (01:40) correct
15%
(01:58)
wrong
based on 204
sessions
History
Date
Time
Result
Not Attempted Yet
Attachment:
two triangles on a line.JPG [ 20.29 KiB | Viewed 9069 times ]
(A) 30°
(B) 35°
(C) 40°
(D) 45°
(E) 50°
This is a relatively straightforward geometry question. For a full discussion of angles & parallel lines, as well as the solution to this question, see:
https://magoosh.com/gmat/2013/angles-and ... -the-gmat/
Experts --- are there any tips you would like to add about parallel lines & angles?
Mike
06 Jun 2013, 12:57
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Step 1: Angle C (also called angle ACB) has to be 40º, because all angles of a triangle must sum 180º
Step 2: Therefore EAD has to be also 40º (i.e. equal to ACB), because that's a property if AD and BC are parallel
Step 3: Finally, angle DEA must be 45º, because (again) all angles of a triangle must sum 180º
Solution D: 45º
Step 2: Therefore EAD has to be also 40º (i.e. equal to ACB), because that's a property if AD and BC are parallel
Step 3: Finally, angle DEA must be 45º, because (again) all angles of a triangle must sum 180º
Solution D: 45º
lchen
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07 Jun 2013, 06:30
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Another way is to see that angle BAF (I'm going to call the leftmost endpoint F because there is no letter for it) is 65 because AD\\BC. Then find DAE by calculating 180-75-65=40. Finally, find AED by calculating 180-40-95=45.
Answer is D
Answer is D
