Difference between revisions of "2012 AMC 10A Problems/Problem 4"
(→Problem 4) |
|||
Line 1: | Line 1: | ||
− | == 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 angle CBD? | ||
Line 9: | Line 9: | ||
<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}} |
Revision as of 22:45, 8 February 2012
Problem
Let and . What is the smallest possible degree measure for angle CBD?
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 |