2008 AMC 12B Problems/Problem 9: Difference between revisions
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==Solution== | ==Solution== | ||
=Trig Solution:= | ===Trig Solution:=== | ||
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> | ||
Revision as of 19:58, 30 November 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
Trig 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.
Other Solution
Define D as the midpoint of AB, and R the center of the circle. R, C, and D are collinear, and since D is the midpoint of AB, 
, and so 
. Since 
, 
, and so 
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 | |
