OA = OB = r = 7.99876785 cos(∠AOC) = = = ∠AOC = 67.97211683° ∠AOB = ∠AOC + ∠BOC = 2(67.97211683°) = 135.9442337° Area of Sector AXBO ∠AOB -------------------- = ------ Area of whole circle 360° Area of Sector AXBO 135.9442337° -------------------- = ------------- 201 360° Area of Sector AXBO = Area of Sector AXBO = Area of Sector AXBO = 75.90219713 Area of circle Segment AXBC = Area of Sector AXBO - Area of ΔABO Now we must find the area of ΔABO It is twice the area of ΔACO We have its height OC = 3 We need its base AC tan(∠AOC) = OC·tan(∠AOC) = AC 3·2.471623109 = AC 7.414869326 = AC Area of ΔACO = = = 11.12230399 Area of ΔABO = 2·Area of ΔACO = 2(11.12230399) = 22.24460798 Area of circle Segment AXBC = Area of Sector AXBO - Area of ΔABO Area of circle Segment AXBC = 75.90219713 - 22.24460798 = 53.65758918 Answer: 53.65758918 round off as your teacher instructed. The numbers here are as far as calculator gives them. ----------------------------------------------------- There is a formula that gives the area of a sector of a circle directly from the central angle, which would have been easier, but I think your teacher expected you to do it the above longer way. But, anyway, that formula is: Area of sector = = the central angle = ∠AOB = 135.9442337° Area of sector = = 53.65758915 Notice that the very last digit differs from the very last digit when calculated above. But you can expect that because a tiny bit of error is made whenever more calculations are made, and there were more calculations made in the above than when substituting directly into the formula. But they are the same when rounded off, even to 7 decimal places. Edwin