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

Revision as of 16:53, 3 July 2013

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 $\frac{5\pi}{3} - \frac{3\pi}{2} = \frac{\pi}{6} \Rightarrow A$ .

Alternative Solution: Since $cos(x) = 0$ , we know that x must equal some multiple of 90. Let us assume x = 90. We want $cos(x+z) = 1/2$ , and by using the property that $cos(x) = cos(180-x)$ , we want x = 60 since $cos(60) = \frac{1}{2}$ . This means that we have $x + z = 120$ , and from this we see that z = 30, or in radians $\frac{\pi}{6}$ .

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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 {{AMC12 box|year=2006|ab=A|num-b=14|num-a=16}}
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 [[Category:Introductory Algebra Problems]]

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

Revision as of 16:53, 3 July 2013

Problem

Solution

See also