Difference between revisions of "1999 AIME Problems/Problem 11"

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== Problem ==
 
== Problem ==
Given that <math>\displaystyle \sum_{k=1}^{35}\sin 5k=\tan \frac mn,</math> where angles are measured in degrees, and <math>\displaystyle m_{}</math> and <math>\displaystyle n_{}</math> are relatively prime positive integers that satisfy <math>\displaystyle \frac mn<90,</math> find <math>\displaystyle m+n.</math>
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Given that <math>\sum_{k=1}^{35}\sin 5k=\tan \frac mn,</math> where angles are measured in degrees, and <math>m_{}</math> and <math>n_{}</math> are relatively prime positive integers that satisfy <math>\frac mn<90,</math> find <math>m+n.</math>
  
 
== Solution ==
 
== Solution ==
  
 
== See also ==
 
== See also ==
* [[1999_AIME_Problems/Problem_10|Previous Problem]]
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{{AIME box|year=1999|num-b=10|num-a=12}}
* [[1999_AIME_Problems/Problem_12|Next Problem]]
 
* [[1999 AIME Problems]]
 

Revision as of 10:36, 11 October 2007

Problem

Given that $\sum_{k=1}^{35}\sin 5k=\tan \frac mn,$ where angles are measured in degrees, and $m_{}$ and $n_{}$ are relatively prime positive integers that satisfy $\frac mn<90,$ find $m+n.$

Solution

See also

1999 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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All AIME Problems and Solutions