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2010 AMC 10B Problem 6

2010 AMC 10B Problem 6AMC 10 Step by Step

Name: Date: September 13, 2026

Circles · difficulty 1 of 5 · about 1 min · position P1-10

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 $\angle CAB$ ?

A)
$20$
B)
$25$
C)
$45$
D)
$50$
E)
$65$

Answer

Key insight

Angle CAB is an inscribed angle on arc CB, while angle COB is the central angle on the same arc, so it is half of 50 degrees.

Solution

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 $\boxed{\textbf{(B)}\ 25}$ .

Why this works

The inscribed angle theorem is the fastest tool whenever a central angle and an inscribed angle share an arc. The fact that $AB$ is a diameter is not even needed for this computation, although it confirms $\angle ACB = 90^\circ$ .

Alternative approach

Triangle $AOC$ is isosceles with $OA = OC$ (radii). Since $A$ , $O$ , $B$ are collinear, $\angle AOC = 180^\circ - 50^\circ = 130^\circ$ , so the two base angles are each $\frac{180^\circ - 130^\circ}{2} = 25^\circ$ .

The trap

Answering 50 by treating the inscribed angle as equal to the central angle, or 65 by using the wrong base angle of the isosceles triangle.

Common mistakes

Answering 50 by treating the inscribed angle as equal to the central angle, or 65 by using the wrong base angle of the isosceles triangle.
Confusing $\angle CAB$ with $\angle ACB$ , which is $90^\circ$ because $AB$ is a diameter.

Techniques

Set up the equation/formula and compute; no special trick needed