ABC are points on a circle, centre O. AO=9cm, OC=9cm and AC=15cm. Find the angle ABC.
Diagrams would be used to help visualise the answer. To find the angle OAC, the cosine rule needs to be used: cosA = (b2 + c2 - a2)/2bc. Therefore, looking at the diagram, (152 + 92 - 92)/2x15x9 = cosA cosA= 5/6, cos-1(5/6) = 33.56 degrees (to 4sf). This is angle OAC. Since AOC is an isoceles triangle, angle OAC and angle OCA are equal, so angle OCA is also 33.56 degrees (to 4sf). Because all angles in a triangle add up to 180 degrees, the final angle AOC can be calculated: 180-(33.56+33.56) = 112.88 degrees. Using circle theorems: the angle at the centre is twice the angle at the circumference. Therefore, 112.88/2 = 56.44 degrees. So angle ABC = 56.44 degrees (to 4sf).
IP
