Difference between revisions of "2010 AMC 10B Problems/Problem 6"

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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==
 
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)
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==Solution 2 (Alcumus)==
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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>.
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<asy>
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import graph;
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unitsize(2 cm);
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pair O, A, B, C;
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O = (0,0);
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A = (-1,0);
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B = (1,0);
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C = dir(50);
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draw(Circle(O,1));
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draw(B--A--C--O);
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label("$A$", A, W);
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label("$B$", B, E);
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label("$C$", C, NE);
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label("$O$", O, S);
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</asy>
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==Video Solution==
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https://youtu.be/I3yihAO87CE
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 +
~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 $O$, $\overline{AB}$ is a diameter and $C$ is a point on the circle with $\angle COB = 50^\circ$. What is the degree measure of $\angle CAB$?

$\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 25 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}\ 65$

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 $O$ is the center, $OC$ and $OA$ are radii and they are congruent. Thus, $\triangle COA$ is an isosceles triangle. Also, note that $\angle COB$ and $\angle COA$ are supplementary, then $\angle COA = 180 - 50 = 130^{\circ}$. Since $\triangle COA$ is isosceles, then $\angle OCA \cong \angle OAC$. They also sum to $50^{\circ}$, so each angle is $\boxed{\textbf{(B)}\ 25}$.


Solution 2 (Alcumus)

Note that $\angle AOC = 180^\circ - 50^\circ = 130^\circ$. Because triangle $AOC$ is isosceles, $\angle CAB = (180^\circ - 130^\circ)/2 = \boxed{25^\circ}$.

[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

2010 AMC 10B (ProblemsAnswer KeyResources)
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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