Difference between revisions of "2012 AMC 10A Problems/Problem 4"

(Problem 4)
(Problem 4)
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<math> \textbf{(A)}\ 0\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 12 </math>
 
<math> \textbf{(A)}\ 0\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 12 </math>
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== Solution ==
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<math>\angle ABD</math> and <math>\angle ABC</math> share ray <math>AB</math>. In order to minimize the value of <math>\angle CBD</math>, <math>D</math> should be located between <math>A</math> and <math>C</math>.
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<math>\angle ABC = \angle ABD + \angle CBD</math>, so <math>\angle CBD = 4</math>. The answer is <math> \qquad\textbf{(C)}</math>

Revision as of 19:30, 8 February 2012

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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