Results 1 to 4 of 4

Math Help - Simple Trig Problem.

  1. #1
    Newbie
    Joined
    Sep 2009
    Posts
    6

    Simple Trig Problem.

    I have a circle. 5 hole equally spaced. A line running from hole 1 to 3. Length of this line is 3.000 inches. What is the diameter of the cirlcle?

    Any help would be appreciated.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by DaveGoad View Post
    I have a circle. 5 hole equally spaced. A line running from hole 1 to 3. Length of this line is 3.000 inches. What is the diameter of the cirlcle?

    Any help would be appreciated.
    Several things are unclear, including: How do the holes relate to the circle? How does the line run from hole 1 to hole 3? What do the holes look like?

    Either explain the question more clearly or (and this is the prefered option) attach a diagram.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,658
    Thanks
    598
    Hello, DaveGoad!

    The word "holes" is confusing.
    Did you mean "points"?


    I have a circle with 5 points equally spaced.
    A line running from point 1 to point 3 has length of 3.000 inches.
    What is the diameter of the cirlcle?
    Code:
          A
          o *
          |\    *
          | \      *   B
        r |  \       o
          |   \
          |    \ 3    *
          |     \     *
        O o 144° \    *
             *    \
             r  *  \ *
                    o
                     C

    The circle has center O.
    Point 1 is A, point 2 is B, point 3 is C.
    Diagonal AC = 3.000 inches.
    Note that: . \angle AOC \,=\,144^o
    The radius is: . r \:=\:OA \:=\:OC

    Law of Cosines: . r^2 + r^2 - 2r^2\cos144^o \:=\:3^2

    . . 2r^2 - 2r^2\cos144^o \:=\:9 \quad\Rightarrow\quad 2r^2(1 - \cos144^o) \:=\:9

    . . 4r^2\left(\frac{1-\cos144^o}{2}\right) \:=\:0 \quad\Rightarrow\quad 4r^2\sin^272^o \:=\:9

    . . r^2 \:=\:\frac{9}{4\sin^272^o} \quad\Rightarrow\quad r \:=\:\frac{3}{2\sin72^o} \:=\:1.577193336


    The diameter is: . d \;=\;2r \;=\;3.154386673 \;\approx\;3.154\text{ inches}

    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Sep 2009
    Posts
    6
    Thank you soroban. Yes I did mean points, the trade im studying for uses the term holes. But the proper term is points, and thank you your soloution is very helpful.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. confusion with simple trig problem
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: August 30th 2011, 11:46 AM
  2. Another Simple Trig Problem
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: September 23rd 2009, 08:39 AM
  3. Simple Statics / Trig problem
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: September 15th 2009, 06:37 AM
  4. Simple Trig Problem Which I Can't Seem To Get!
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: April 22nd 2009, 03:36 AM
  5. Very simple trig problem
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: June 6th 2007, 09:17 PM

Search Tags


/mathhelpforum @mathhelpforum