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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