1956 AMC 12 Problem 10

Attempt Problem 10 of the 1956 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1956 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

10.

A circle of radius

10

inches has its center at the vertex

C

of an equilateral triangle

ABC

and passes through the other two vertices. The side

AC

extended through

C

intersects the circle at

D.

The number of degrees of angle

ADB

is:

15

30

60

90

120

The radii

CA

and

CB

form a

60^\circ

central angle

Angle

ADB

is an inscribed angle intercepting the minor arc

AB

Since

ABC

is equilateral, the central angle

\angle ACB

60^\circ.

The inscribed angle

\angle ADB

intercepts the same minor arc

AB,

so its measure is half the central angle:

\angle ADB=\frac{60^\circ}{2}=30^\circ.

Thus, the correct answer is B.

Problem 10 in other years