Given that AB:BC = 5:6 and angle ABC = 48 degrees, we can find the other angles in triangle ABC.

To find angle BAC:
Let x be the measure of angle BAC.
Since angle ABC + angle BAC + angle ACB = 180 degrees in a triangle,
48 + x + 180 - 48 - x = 180
x = 180 - 48 = 132 degrees

Therefore, angle BAC = 132 degrees.

To find angle AOC:
Since AB:BC = 5:6, we can determine the ratio of the sides of triangle ABC.
Let the lengths of AB and BC be 5x and 6x respectively.
Now, let's use the Law of Cosines to find angle AOC:
cos(AOC) = (5x^2 + 6x^2 - AC^2) / (2 5x 6x)
cos(AOC) = (61x^2 - AC^2) / (60x^2)
Since angle AOC is an exterior angle, the sum of the interior angles A and C will equal angle AOC.
Therefore, cos(AOC) = cos(ACB + BAC) = cos(48 + 132) = cos(180) = -1
-1 = (61x^2 - AC^2) / (60x^2)
61x^2 - AC^2 = -60x^2

Solving for AC:
AC^2 = 121x^2
AC = 11x

Therefore, angle AOC = 180 - 48 = 132 degrees.