1956 AHSME Problems/Problem 10

Problem

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:

$\text{(A)} 15 \quad \text{(B)}30 \quad \text{(C)}60 \quad \text{(D)}90 \quad \text{(E)}120$

Solution

$[asy] import olympiad; draw(circle((0,0),10)); dot((0,0)); label("C", (1,-1)); dot((5,5sqrt(3))); dot((-5,-5sqrt(3))); draw((-5,-5sqrt(3))--(5,5sqrt(3))); label("A",(6,5sqrt(3)+1)); label("D",(-6,-5sqrt(3)-1)); draw((0,0)--(10,0)); label("10",(1.5,2.5sqrt(3)+1)); dot((10,0)); label("B",(11,-1)); draw((5,5sqrt(3))--(10,0)); draw(anglemark((10,0),(0,0),(5,5sqrt(3)),60)); label( "60°", (2.5,1.25)); draw((-5,-5sqrt(3))--(10,0)); draw(anglemark((10,0),(-5,-5sqrt(3)),(5,5sqrt(3)),60)); label("?",(-2.5,-5sqrt(3)+2.5)); [/asy]$

$ABC$ is an equilateral triangle, so ∠ $C$ must be $60$ °. Since $D$ is on the circle and ∠ $ADB$ contains arc $AB$ , we know that ∠ $D$ is $30$ ° $\implies \fbox{B}$ .

1956 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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1956 AHSME Problems/Problem 10

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