Posted by T on Saturday, May 24, 2008 at 10:43pm.
Area of triangle = side1*side2*sin(angle between them)/2
so 48 = 1/2(12)(9)sin ß
sin ß = 48/54
ß = 62.7º or 117.26º
If the angle is 62.7
Using the cosine law I found the side across from the 62.7º to be 11.226 and then using the sine law once more found the angle across from the side of 9 to be 45.44º surely making all three angles acute.
So the largest angle in the triangle must be in the second case, where the angle was 117.7º
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