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

Latest revision as of 12:56, 1 July 2023

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

~Education, the Study of Everything

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

@@ Line 1: / Line 1: @@
-== Problem 4 ==
+== Problem ==
-Let <math>\angle ABC = 24</math>° and <math>\angle ABD = 20</math>°. What is the smallest possible degree measure for <math>\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> \qquad\textbf{(A)}\ 0\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 12 $
+<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 ==
+<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> \boxed{\textbf{(C)}\ 4}</math>
+==Video Solution (CREATIVE THINKING)==
+https://youtu.be/ehawVnD0wc4
+~Education, the Study of Everything
+== See Also ==
+{{AMC10 box|year=2012|ab=A|num-b=3|num-a=5}}
+{{MAA Notice}}