Difference between revisions of "2007 AMC 12B Problems/Problem 3"
(→Problem) |
(→Alternative Solution) |
||
| (6 intermediate revisions by 4 users not shown) | |||
| Line 6: | Line 6: | ||
<math>\mathrm {(A)} 35 \qquad \mathrm {(B)} 40 \qquad \mathrm {(C)} 45 \qquad \mathrm {(D)} 50 \qquad \mathrm {(E)} 60</math> | <math>\mathrm {(A)} 35 \qquad \mathrm {(B)} 40 \qquad \mathrm {(C)} 45 \qquad \mathrm {(D)} 50 \qquad \mathrm {(E)} 60</math> | ||
==Solution== | ==Solution== | ||
| − | Since triangles <math>ABO</math> and <math>BOC</math> are isosceles, <math>\angle ABO=20^o</math> and <math>\angle OBC=30^o</math>. Therefore, <math>\angle ABC=50^o</math>, or | + | Since triangles <math>ABO</math> and <math>BOC</math> are isosceles, <math>\angle ABO=20^o</math> and <math>\angle OBC=30^o</math>. Therefore, <math>\angle ABC=50^o</math>, or <math>\mathrm{(D)}</math>. |
| + | |||
| + | ==Alternative Solution== | ||
| + | <math>\angle AOC = 100^{\circ} \implies \angle ABC =\frac{\angle AOC}{2} =50^{ \circ},</math> or <math>\mathrm{(D)}.</math> | ||
==See Also== | ==See Also== | ||
Latest revision as of 09:39, 27 February 2022
Problem
The point 
is the center of the circle circumscribed about triangle 
, with 
and 
, as shown. What is the degree measure of 
?
Solution
Since triangles 
and 
are isosceles, 
and 
. Therefore, 
, or 
.
Alternative Solution
or 
See Also
| 2007 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 2 |
Followed by Problem 4 |
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
