The answer depends on what c represents and other information as well.
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The length of arc ACB is 57.2.
All you've told us is that 40 cm is less than 1/2 of the circumference. With that information, all we know is that the circumference is more than 80 cm. We could calculate it if we knew what the angle is at the center of circle between the two radii (radiuses) that go to the ends of arc AB. We're guessing that it's there in your book, but you forgot to include it when you decided to ask us to do your homework problem for you.
Semicircle
If angle ACB is the right angle then ab is the hypotenuse. Then, (ab)2 = 62 + 92 = 36 + 81 = 117 ab = √117 = 10.8 (3 sf) If angle BAC is the right angle then ab is one leg of a right angled triangle with bc the hypotenuse. 92 = 62 + (ab)2 : (ab)2 = 92 - 62 = 81 - 36 = 45 ab = √45 = 6.71 (3 sf)
The picture that you see printed next to that question on the assignment sheet is necessary in order to answer it. Without the picture, I have no idea how angle-1, arc-ab, and arc-ag are connected .