Difference between revisions of "2006 AMC 12A Problems/Problem 15"

Revision as of 18:20, 2 February 2007

Problem

Suppose $\cos x=0$ and $\cos (x+z)=1/2$ . What is the smallest possible positive value of $z$ ?

$\mathrm{(A) \ } \frac{\pi}{6}\qquad \mathrm{(B) \ } \frac{\pi}{3}\qquad \mathrm{(C) \ } \frac{\pi}{2}\qquad \mathrm{(D) \ } \frac{5\pi}{6} \quad \mathrm{(E) \ } \frac{7\pi}{6}$

Solution

For $\cos x = 0$ , x must be in the form of $\frac{\pi}{2} + \pi n$ , where $n$ denotes any integer.
For $\cos (x+z) = 1 / 2$ , $x + z = \frac{\pi}{3} +2\pi n, \frac{5\pi}{3} + 2\pi n$ .

The smallest possible value of $z$ will be that of $\frac{5\pi}{3} - \frac{3\pi}{2} = \frac{\pi}{6} \Rightarrow A$ .

2006 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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 12 Problems and Solutions

@@ Line 14: / Line 14: @@
 * [[2006 AMC 12A Problems]]
-{{AMC box|year=2006|n=12A|num-b=14|num-a=16}}
+{{AMC12 box|year=2006|ab=A|num-b=14|num-a=16}}
 [[Category:Introductory Algebra Problems]]

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Difference between revisions of "2006 AMC 12A Problems/Problem 15"

Revision as of 18:20, 2 February 2007

Problem

Solution

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