Difference between revisions of "2010 AMC 10B Problems/Problem 6"
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==Solution 2== | ==Solution 2== | ||
| − | An inscribed angle is always half its central angle, so therefore, half of 50 is 25, or B. | + | An inscribed angle is always half its central angle, so therefore, half of 50 is 25, or B. We can see this when we graph the problem. |
(Solution by Flamedragon) | (Solution by Flamedragon) | ||
Revision as of 21:11, 9 February 2014
Contents
Problem
A circle is centered at 
, $\overbar{AB}$ (Error compiling LaTeX. Unknown error_msg) is a diameter and 
is a point on the circle with 
. What is the degree measure of 
?
Solution 1
Assuming we do not already know an inscribed angle is always half of its central angle, we will try a different approach. Since is the center, 
and 
are radii and they are congruent. Thus, 
is an isosceles triangle. Also, note that 
and 
are supplementary, then 
. Since is isosceles, then 
. They also sum to 
, so each angle is 
.
Solution 2
An inscribed angle is always half its central angle, so therefore, half of 50 is 25, or B. We can see this when we graph the problem.
(Solution by Flamedragon)
See Also
| 2010 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 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 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
