Problem One
Any angle drawn from the end points of a diameter and whose vertex is on the circumference of a circle is a right angle. <ACB is a right angle. Therefore you can use Pythagoras.
Formula and givens
a = 12
b = 5
a^2 + b^2 = c^2
c^2 = 12^2 + 5^2
c^2 = 144 + 25
c^2 = 169 Take the sqrt of both sides.
sqrt(c^2)=sqrt(169)
c = 13
The radius = d/2 = 13/2 = 6.5 cm
Problem 2
Find <AOC. Begin by drawing OC
The measure of arc AC = 40 degrees, Because AOC is a central angle, it is also 40 degrees. Triangle AOC is isosceles. That comes from OA and OC both being radii. If angle AOC is 40o then <AOC + OAC + OCA = 180
Call <OCA = <OAC = x
40 + x + x = 180
40 + 2x = 180 Subtract 40 from both sides
2x = 180 - 40
2x = 140
x = 70
< OAC = 70
