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

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<!-- explanation needed-->The smallest possible value of <math>z</math> will be that of <math>\frac{5\pi}{3} - \frac{3\pi}{2} = \frac{\pi}{6} \Rightarrow \mathrm{(A)}</math>.
 
<!-- explanation needed-->The smallest possible value of <math>z</math> will be that of <math>\frac{5\pi}{3} - \frac{3\pi}{2} = \frac{\pi}{6} \Rightarrow \mathrm{(A)}</math>.
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=== Solution 2 ===
 
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Since <math>cos(x) = 0</math>, we know that x must equal some multiple of 90. Let us assume x = 90. We want <math>cos(x+z) = 1/2</math>, and by using the property that <math>cos(x) = cos(180-x)</math>, we want x = 60 since <math>cos(60) = \frac{1}{2}</math>. This means that we have <math>x + z = 120</math>, and from this we see that z = 30, or in radians <math>\frac{\pi}{6}</math>.
 
  
 
== See also ==
 
== See also ==

Revision as of 15:54, 12 October 2016

Problem

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

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

Solutions

Solution 1

  • 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 $\frac{5\pi}{3} - \frac{3\pi}{2} = \frac{\pi}{6} \Rightarrow \mathrm{(A)}$.

See also

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

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