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Series and sum | AIME I, 1999 | Question 11

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1999 based on Series and sum.

Series and sum - AIME I, 1999

given that \(\displaystyle\sum_{k=1}^{35}sin5k=tan\frac{m}{n}\) where angles are measured in degrees, m and n are relatively prime positive integer that satisfy \(\frac{m}{n} \lt 90\), find m+n.

  • is 107
  • is 177
  • is 840
  • cannot be determined from the given information

Key Concepts



Side Length

Check the Answer

Answer: is 177.

AIME I, 2009, Question 5

Plane Trigonometry by Loney

Try with Hints

First hint


Second Hint


Final Step

\(=\frac{1-cos(175)}{sin175}\)=\(tan\frac{175}{2}\) then m+n=175+2=177.

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