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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 -

    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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