2010 AMC 10B Problems/Problem 6

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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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