2012 AMC 10A Problems/Problem 4

Problem

Let $\angle ABC = 24^\circ$ and $\angle ABD = 20^\circ$ . What is the smallest possible degree measure for $\angle CBD$ ?

$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 12$

$\angle ABD$ and $\angle ABC$ share ray $AB$ . In order to minimize the value of $\angle CBD$ , $D$ should be located between $A$ and $C$ .

$\angle ABC = \angle ABD + \angle CBD$ , so $\angle CBD = 4$ . The answer is $\boxed{\textbf{(C)}\ 4}$

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2012 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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