#2**+1 **

Call O the center of the circle.....draw radius OC perpendicular to AB, and connect OB and OA

In triangle OBC...angle OBC = 30° and angle BOC = 60°

So....BC = 6√3 cm

And in triangle OAC = 15“ and angle COA = 75°

Using the law of sines

CA / sin 75 = OC / sin 15

CA / sin (30 + 45) = 6 / sin (45 -30)

CA = 6 sin (30 + 45) / sin (45 - 30)

CA = 6 [ sin 30cos 45 + sin 45cos30 ] / [ sin 45cos 30 - sin 30 cos 45 ]

CA = 6 [ 1/2 * √2/2 + √2/2 * √3/2 ] / [ √2/2 * √3 / 2 * 1/2 * √2/2 ]

CA = 6 [ √2/4 + √6/4 ] / [ √6 / 4 - √2/4]

CA = 6 [ √6 + √2 ] / [√6 - √2]

CA = 6 [ 6 + 2√12 + 2 ] / [6 - 2]

CA = 6 [8 + 4√3 ] / [ 4]

CA = (3/2) [ 8 + 4√3] = 3 [ 4 + 2√3] = 12 + 6√3 cm

So....AB = BC + CA = 6√3 + 12 + 6√3 = 12 + 12√3 cm

CPhill Oct 14, 2018