Difference between revisions of "2010 AMC 10B Problems/Problem 6"
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==Solution 1== | ==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>. | ||
+ | |||
+ | |||
+ | ==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== | ||
+ | https://youtu.be/I3yihAO87CE | ||
+ | |||
+ | ~IceMatrix | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2010|ab=B|num-b=5|num-a=7}} | {{AMC10 box|year=2010|ab=B|num-b=5|num-a=7}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest 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, .
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 |
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