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 ==
-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>
+[[Category: Introductory Geometry Problems]]
 == Solution ==
@@ Line 10: / Line 11: @@
 <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}}