Difference between revisions of "2008 AMC 12B Problems/Problem 9"
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The [[Trigonometric_identities#Half_Angle_Identities | half-angle formula]] says that | The [[Trigonometric_identities#Half_Angle_Identities | half-angle formula]] says that | ||
− | <math>\cos(\alpha/2) = \frac{\sqrt{1+\cos(\alpha)}}{2} = \sqrt{\frac{32/25}{2}} = \sqrt{\frac{16}{25}} = \frac{4}{5}</math>. The law of cosines tells us <math>AC = \sqrt{5^2+5^2-2 | + | <math>\cos(\alpha/2) = \frac{\sqrt{1+\cos(\alpha)}}{2} = \sqrt{\frac{32/25}{2}} = \sqrt{\frac{16}{25}} = \frac{4}{5}</math>. The law of cosines tells us <math>AC = \sqrt{5^2+5^2-2\cdot 5\cdot 5\cdot \frac{4}{5}} = \sqrt{50-50\frac{4}{5}} = \sqrt{10}</math>, which is answer choice <math>\boxed{\text{A}}</math>. |
===Solution 2=== | ===Solution 2=== |
Revision as of 23:23, 14 August 2011
Contents
[hide]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 ?
Solutions
Solution 1
Let be the angle that subtends the arc . By the law of cosines, implies .
The half-angle formula says that . The law of cosines tells us , which is answer choice .
Solution 2
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Figure 1 |
Define as the midpoint of line segment , and the center of the circle. Then , , and are collinear, and since is the midpoint of , and so . Since , , and so .
See Also
2008 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 8 |
Followed by Problem 10 |
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 |