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

Revision as of 19:30, 8 February 2012

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 $\qquad\textbf{(C)}$

@@ Line 4: / Line 4: @@
 <math> \textbf{(A)}\ 0\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 12 </math>
+== Solution ==
+<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>.
+<math>\angle ABC = \angle ABD + \angle CBD</math>, so <math>\angle CBD = 4</math>. The answer is <math> \qquad\textbf{(C)}</math>