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2012 AMC 10A Problems/Problem 4

Revision as of 20:30, 8 February 2012 by Flamewire (talk | contribs) (→‎Problem 4)

Problem 4

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

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

Solution

$\angle ABD$ 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 $\qquad\textbf{(C)}$

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