2010 AMC 10B Problem 6

Draw the picture: $A$ and $B$ at the two ends of a diameter, $O$ in the middle, and $C$ on the circle so that $\angle COB = 50^\circ$.

1. Look at which part of the circle each angle "sees". Both $\angle COB$ and $\angle CAB$ open onto the same arc, the arc from $C$ to $B$.
2. $\angle COB$ has its corner at the center, so the arc $CB$ measures $50^\circ$.
3. $\angle CAB$ has its corner $A$ on the circle. An angle with its corner on the circle is always half of the arc it opens onto. So
$$\angle CAB = \frac{50^\circ}{2} = 25^\circ.$$

The answer is $\boxed{\textbf{(B)}\ 25}$.

This is the inscribed angle rule: an angle with its corner on the circle is half the angle at the center that opens onto the same arc. Whenever a problem gives you an angle at the center and asks about an angle at a point on the circle (or the other way round), check whether they share an arc.

If you do not remember the rule, use an isosceles triangle.

1. $OA$ and $OC$ are both radii, so triangle $AOC$ has two equal sides, and its two base angles at $A$ and $C$ are equal.
2. $A$, $O$, $B$ lie on one straight line, so $\angle AOC = 180^\circ - 50^\circ = 130^\circ$.
3. The two base angles share what is left: $\frac{180^\circ - 130^\circ}{2} = 25^\circ$. The base angle at $A$ is exactly $\angle CAB$.

Both $\angle COB$ and $\angle CAB$ open onto the same arc $CB$. The first has its vertex at the center, so it is a central angle equal to the arc, $50^\circ$. The second has its vertex $A$ on the circle, so it is an inscribed angle, and an inscribed angle is half the arc it cuts:
$$\angle CAB = \frac{1}{2}\cdot 50^\circ = 25^\circ.$$

The answer is $\boxed{\textbf{(B)}\ 25}$.