FG is a chord (having length 6) of a circle with center O having radius 5.
E is a point on the circle closer to F than G such that there is just one chord EH which is bisected by FG.
Find sin ∠EOF.
The half chord (half of FG) has length 3. As the radius of the circle is 5, the chord is 4 units distant from O.
Consider the perpendicular bisector of FG: it goes through the center O.
Any chord, including EH, that's bisected by FG, has endpoints that lie on chords parallel to FG. If E lies on the chord farther from O than is FG, then the other parallel chord will intersect the circle in two points, so E must be "below" FG, that is, on a chord parallel to FG that's closer to O. And, in fact, the upper (more distant from O) parallel chord, in order to make the end point H unique, must be at the "top" of the circle (tangent to the circle on the perpendicular bisector of FG).
Under these conditions E is 4 units away from the perpendicular bisector of FG; its chord is 3 units away, as its own distance from O is 5. F, as mentioned, is 3 units away from the perpendicular bisector and FG is 4 units from O.
A diagram shows the angle EOF is atan(4/3) - atan(3/4).
A calculator shows the sine of this angle to be 0.28 exactly, or 7/25.
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Posted by Charlie
on 2016-01-11 15:13:50 |
solution
