Difference between revisions of "2008 AMC 12B Problems/Problem 9"
(New page: ==Problem 9== Points <math>A</math> and <math>B</math> are on a circle of radius <math>5</math> and <math>AB = 6</math>. Point <math>C</math> is the midpoint of the minor arc <math>AB</mat...) |
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| Line 7: | Line 7: | ||
Let <math>\alpha</math> be the angle that subtends the arc AB. By the law of cosines, | Let <math>\alpha</math> be the angle that subtends the arc AB. By the law of cosines, | ||
<math>6^2=5^2+5^2-2*5*5cos(\alpha)</math> | <math>6^2=5^2+5^2-2*5*5cos(\alpha)</math> | ||
| + | |||
<math>\alpha = cos^{-1}(7/25)</math> | <math>\alpha = cos^{-1}(7/25)</math> | ||
Revision as of 03:18, 2 March 2008
Problem 9
Points 
and 
are on a circle of radius 
and 
. Point 
is the midpoint of the minor arc 
. What is the length of the line segment 
?
Solution
Let 
be the angle that subtends the arc AB. By the law of cosines,
The half-angle formula says that
, which is answer choice A.
See Also
| 2008 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 7 |
Followed by Problem 9 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
