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

(Created page with 'Assuming the reader is not readily capable to understand how <math>\angle ACB</math> will always be right, the I will continue with an easily understandable solution. Since <math…')
 
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Assuming the reader is not readily capable to understand how <math>\angle ACB</math> will always be right, the I will continue with an easily understandable solution. Since <math>O</math> is the center, <math>OC, OB, \text{ and } OA</math> are all radii, they are congruent. Thus, <math>\triangle COB</math> and <math>\triangle COA</math> are isosceles triangles. Also, note that <math>\angle COB</math> and <math>\angle COA</math> are supplementary, then <math>\angle COA = 180 - 50 = 130 \degrees</math>. Since <math>\triangle COA</math> is isosceles, then <math>\angle OCA \cong \angle OAC</math>. They also sum to <math>50 \degrees</math>, so each angle is <math>\boxed{\mathrm{(B)} 25 \degrees}</math>.
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==Problem==
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A circle is centered at <math>O</math>, <math>\overline{AB}</math> is a diameter and <math>C</math> is a point on the circle with <math>\angle COB = 50^\circ</math>. What is the degree measure of <math>\angle CAB</math>?
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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>
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==Solution 1==
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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>.
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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==
 +
https://youtu.be/wqRMJBoSm_A
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 +
~Education, the Study of Everything
 +
 
 +
==Video Solution==
 +
https://youtu.be/I3yihAO87CE
 +
 
 +
~IceMatrix
 +
 
 +
==See Also==
 +
{{AMC10 box|year=2010|ab=B|num-b=5|num-a=7}}
 +
{{MAA Notice}}

Latest revision as of 16:00, 1 August 2022

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/wqRMJBoSm_A

~Education, the Study of Everything

Video Solution

https://youtu.be/I3yihAO87CE

~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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