We are given a semicircle with diameter AB = 100. Points C and D are on the semicircle such that AC = 28 and BD = 60. We are asked to determine CD.

Here is one approach. (See diagram below.)

Let the center of the semicircle be O.

1) Consider `Delta ABC` . This is a right triangle (an angle inscribed in a semicircle is a right angle). The measure of `angle ABC` is `sin^(-1)(28/100) ~~ 16.26^@` . Then the measure of arc AC is approximately `32.52^@` .

2) Let E be the intersection of the radius drawn ` _|_ ` to the chord BD. `Delta BEO` is a 3-4-5 right triangle. (BE is 30, since the radius bisects t chord, BO is 50, and BE is 40 from the Pythagorean theorem.)

Then `m angle BOD = 2m angle BOE =2tan^(-1)(3/4)~~73.74^@` . Thus the measure of arc BD is also `73.74^@` since an arc has the same measure as its central angle.

3) Arc CD has measure `180-(73.74^@+16.26^@)~~73.74^@` . Since arc CD and arc BD have the same measure, their chords are the same length.

CD = BD = 60.

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