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The problem says it's a rectangle. A rectangle's diagonals bisect each other and are equal in length, so if you draw a rectangle and draw in the diagonals, you'll get 4 isosceles triangles. Both statements give one angle of the two angles between the base and the equal sides, so you can calculate the third angle from both statements and get to AEB.
+1 for D. Use the property that diagonals of a rectangle bisect one another. Both statements are sufficient individually. Hence option D.
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No Bunuel I could not relate the picture with your solution given earlier. Could you please explain it with picture how \(\angle AEB = 180 - 2*\angle EDC = 150^o\)?
No Bunuel I could not relate the picture with your solution given earlier. Could you please explain it with picture how \(\angle AEB = 180 - 2*\angle EDC = 150^o\)?
For (1). First of all, \(\angle AEB = \angle DEC\). Next, DEC is isosceles, thus \(\angle EDC = \angle ECD=15^o\). The sum of the angles in a triangle is 180 degrees:
\(\angle EDC + \angle ECD + \angle DEC = 180^o\);
\(\angle EDC + \angle EDC + \angle AEB = 180^o\);
\(\angle AEB = 180 - 2*\angle EDC = 150^o\).
For (2). First of all, \(\angle BAC + \angle ACB= 90^o\) --> \(\angle BAC =90^o-\angle ACB\). Next, AEB is isosceles, thus \(\angle BAC = \angle ABE\). The sum of the angles in a triangle is 180 degrees. \(\angle BAC + \angle ABE + \angle AEB = 180^o\);
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73% (01:09) correct 
