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? | + | 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> | ||
+ | [[Category: Introductory Geometry Problems]] | ||
== Solution == | == 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>. | <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> \ | + | <math>\angle ABC = \angle ABD + \angle CBD</math>, so <math>\angle CBD = 4</math>. The answer is <math> \boxed{\textbf{(C)}\ 4}</math> |
+ | |||
+ | == See Also == | ||
+ | |||
+ | {{AMC10 box|year=2012|ab=A|num-b=3|num-a=5}} | ||
+ | {{MAA Notice}} |
Latest revision as of 15:03, 13 July 2021
Problem
Let and . What is the smallest possible degree measure for ?
Solution
and share ray . In order to minimize the value of , should be located between and .
, so . The answer is
See Also
2012 AMC 10A (Problems • Answer Key • Resources) | ||
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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