Difference between revisions of "2010 AMC 10B Problems/Problem 6"
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<math>\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 25 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}\ 65</math> | <math>\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 25 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}\ 65</math> | ||
| − | ==Solution== | + | ==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 <math>O</math> is the center, <math>OC</math> and <math>OA</math> are radii and they are congruent. Thus, <math>\triangle COA</math> is an isosceles triangle. Also, note that <math>\angle COB</math> and <math>\angle COA</math> are supplementary, then <math>\angle COA = 180 - 50 = 130^{\circ}</math>. Since <math>\triangle COA</math> is isosceles, then <math>\angle OCA \cong \angle OAC</math>. They also sum to <math>50^{\circ}</math>, so each angle is <math>\boxed{\textbf{(B)}\ 25}</math>. | Assuming we do not already know an inscribed angle is always half of its central angle, we will try a different approach. Since <math>O</math> is the center, <math>OC</math> and <math>OA</math> are radii and they are congruent. Thus, <math>\triangle COA</math> is an isosceles triangle. Also, note that <math>\angle COB</math> and <math>\angle COA</math> are supplementary, then <math>\angle COA = 180 - 50 = 130^{\circ}</math>. Since <math>\triangle COA</math> is isosceles, then <math>\angle OCA \cong \angle OAC</math>. They also sum to <math>50^{\circ}</math>, so each angle is <math>\boxed{\textbf{(B)}\ 25}</math>. | ||
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| + | ==Solution 2 (Alcumus)== | ||
| + | Note that <math>\angle AOC = 180^\circ - 50^\circ = 130^\circ</math>. Because triangle <math>AOC</math> is isosceles, <math>\angle CAB = (180^\circ - 130^\circ)/2 = \boxed{25^\circ}</math>. | ||
| + | |||
| + | <asy> | ||
| + | import graph; | ||
| + | |||
| + | unitsize(2 cm); | ||
| + | |||
| + | pair O, A, B, C; | ||
| + | |||
| + | O = (0,0); | ||
| + | A = (-1,0); | ||
| + | B = (1,0); | ||
| + | C = dir(50); | ||
| + | |||
| + | draw(Circle(O,1)); | ||
| + | draw(B--A--C--O); | ||
| + | |||
| + | label("$A$", A, W); | ||
| + | label("$B$", B, E); | ||
| + | label("$C$", C, NE); | ||
| + | label("$O$", O, S); | ||
| + | </asy> | ||
==Video Solution== | ==Video Solution== | ||
Revision as of 15:40, 13 March 2021
Problem
A circle is centered at 
, 
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 (Alcumus)
Note that 
. Because triangle 
is isosceles, 
.
![[asy] import graph; unitsize(2 cm); pair O, A, B, C; O = (0,0); A = (-1,0); B = (1,0); C = dir(50); draw(Circle(O,1)); draw(B--A--C--O); label("$A$", A, W); label("$B$", B, E); label("$C$", C, NE); label("$O$", O, S); [/asy]](http://latex.artofproblemsolving.com/a/a/b/aab4fe9f0a401fcda884f1328c442a93c5cf59bb.png)
Video Solution
~IceMatrix
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. 
