by tim Thu Jul 08, 2010 1:36 pm
If the area is to be 30, considering 13 as the base the height has to be 60/13, or a little less than 5. Mentally remove segment BC from the drawing, and you'll see that AB is free to swing around point A. We can get different heights depending on how we position AB, but in particular you should note that there are two positions we can place AB that will give us a height of 60/13 - one produces an acute angle at A and the other produces an obtuse angle. Thus there are exactly two distinct values BC can take on that fit the constraints of the problem..
Tim Sanders
Manhattan GMAT Instructor
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