2006 AMC 12A Problems/Problem 15
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 |
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All AMC 12 Problems and Solutions |
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