Posted by Anonymous on Saturday, March 2, 2013 at 2:02pm.
Let ABC be a right triangle with ACB =90, AC =6 , and BC =2. E is the midpoint of AC , and F is the midpoint of AB . If CF and BE intersect at G , then cos(CGB),
in simplest radical form, is ((k square root w)/f) where k , w , and f are positive integers. Find the value of k+w+ f.

geometry  Steve, Saturday, March 2, 2013 at 7:38pm
Since E and F are midpoints, EF ║ CB, and has length CB/2 = 1
CE = AC/2 = 3
Label angles
CGB = a
EBC = b
FCB = c
so, intriangle CGB, a+b+c = 180 degrees
we want
cos a = cos *180  (b+c))
= cos(b+c)
= sinb sinc  cosb cosc
Looking at the right triangles,
EB = √13 and FC = √10, so we have
cos a = 3/√13 * 3/√10  2/√13 * 1/√10
= 9/√130  2/√130
= 7/√130
= 7√130 / 130
k+w+f = 7+130+130 = 267
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