Difference between revisions of "2006 AMC 12A Problems/Problem 15"
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<math> \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}</math> | <math> \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}</math> | ||
− | == Solution == | + | == Solutions == |
+ | == Solution 1 == | ||
*For <math>\cos x = 0</math>, x must be in the form of <math>\frac{\pi}{2} + \pi n</math>, where <math>n</math> denotes any [[integer]]. | *For <math>\cos x = 0</math>, x must be in the form of <math>\frac{\pi}{2} + \pi n</math>, where <math>n</math> denotes any [[integer]]. | ||
*For <math>\cos (x+z) = 1 / 2</math>, <math>x + z = \frac{\pi}{3} +2\pi n, \frac{5\pi}{3} + 2\pi n</math>. | *For <math>\cos (x+z) = 1 / 2</math>, <math>x + z = \frac{\pi}{3} +2\pi n, \frac{5\pi}{3} + 2\pi n</math>. | ||
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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>. | ||
− | + | == Solution 2 == | |
+ | 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 23:17, 5 February 2016
Problem
Suppose and
. What is the smallest possible positive value of
?
Solutions
Solution 1
- For
, x must be in the form of
, where
denotes any integer.
- For
,
.
The smallest possible value of will be that of
.
Solution 2
Since , we know that x must equal some multiple of 90. Let us assume x = 90. We want
, and by using the property that
, we want x = 60 since
. This means that we have
, and from this we see that z = 30, or in radians
.
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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