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

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== Problem ==
 
== Problem ==
  
Let <math>\angle ABC = 24^\circ </math> and <math>\angle ABD = 20^\circ </math>. What is the smallest possible degree measure for angle CBD?
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Let <math>\angle ABC = 24^\circ </math> and <math>\angle ABD = 20^\circ </math>. What is the smallest possible degree measure for <math>\angle CBD</math>?
  
 
<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>

Latest revision as of 15:03, 13 July 2021

Problem

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

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

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 $\boxed{\textbf{(C)}\ 4}$

See Also

2012 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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