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July 31, 2014

July 31, 2014

Posted by **T** on Saturday, May 24, 2008 at 10:43pm.

- MATH -
**Reiny**, Saturday, May 24, 2008 at 11:51pmArea 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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