AB is a diameter of a circle centered at O. C is a point on the circle such that angle BOC is 60°. If the diameter of the circle is 5 inches, the length of chord AC, expressed in inches, is:

1966 AMC 12 Problem 6

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AB

is a diameter of a circle centered at

O.

C

is a point on the circle such that angle

BOC

60^\circ.

If the diameter of the circle is

5

inches, the length of chord

AC,

expressed in inches, is:

3

\dfrac{5\sqrt2}{2}

\dfrac{5\sqrt3}{2}

3\sqrt3

none of these

Because

AB

is a diameter,

\angle AOC=120^\circ

Split isosceles triangle

AOC

into two

30

60

90

triangles

The radius is

\frac{5}{2},

and

\angle AOC=180^\circ-60^\circ=120^\circ.

A chord subtending

120^\circ

has length

2\left(\frac52\right)\sin60^\circ=\frac{5\sqrt3}{2}.

Therefore, the correct answer is C.