Difference between revisions of "2006 AMC 12A Problems/Problem 15"
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| + | Alternative Solution: 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(6) = \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>. | ||
Revision as of 18:09, 17 February 2010
Problem
Suppose 
and 
. What is the smallest possible positive value of 
?
Solution
- For
, x must be in the form of 
, where 
denotes any integer. - For
, 
.
, x must be in the form of 
, where 
denotes any 
, 
.The smallest possible value of will be that of 
.
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 | |
Alternative Solution: 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 
.
