Results 1 to 7 of 7

Math Help - Trigonometry three dimensional question

  1. #1
    Newbie
    Joined
    Jun 2006
    From
    NSW Australia
    Posts
    12

    Trigonometry three dimensional question

    I need help to solve:
    A cylinder with radius 4 cm and perpendicular height 15 cm is tilted so that it will just fit inside a 12 cm high box. At what angle must it be tilted?

    Answer given at the back of the text book is 16 degrees 15 minutes.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by lalji View Post
    I need help to solve:
    A cylinder with radius 4 cm and perpendicular height 15 cm is tilted so that it will just fit inside a 12 cm high box. At what angle must it be tilted?

    Answer given at the back of the text book is 16 degrees 15 minutes.
    Maybe it is the three dimensions that are screwing you up...I say think it more two dimensionally to start...and here is a big hint...PERPENDICULAR height...and perp. is synonomous with what?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,

    Take a look at this picture. OPQR represents the 12cm high box. ABCD is a view from the side of the cylinder.

    According to the text,
    RQ=OP=12 cm

    AB=CD=2 \cdot 4=8 cm

    AC=BD=15 cm

    And you're looking for angle DAR.

    Is it more helpful ?
    Attached Thumbnails Attached Thumbnails Trigonometry three dimensional question-cylinder.jpg  
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,826
    Thanks
    714
    Hello, lalji!

    I don't agree with their answer . . .


    A cylinder with radius 4 cm and perpendicular height 15 cm
    is tilted so that it will just fit inside a 12 cm high box.
    At what angle must it be tilted?

    Answer given: 16 degrees 15 minutes.
    Code:
          P       A         S
        _ * - - - * - - - - *
        : |      *   *      |
        : |     *       *D  |
        : |    *       *    |
      12-x|   * 15    *     |
        : |  *       *      |
        : | *       *       |
        : |*       *        |
        -B*   8   *         |
        x |  *   *          |
        - * - - * - - - - - *
          Q     C           R

    The cylinder is ABCD\!: CB = 8,\;AB = 15,\;PQ = 12

    The box is PQRS\!:\;PQ = 12

    Let x = BQ
    Let \theta = \angle BCQ
    . . Note that \angle ABP = \theta

    In right triangle BCQ\!:\;QC = \sqrt{64-x^2}
    . . \cos\theta \,=\,\frac{\sqrt{64-x^2}}{8}\;\;{\color{blue}[1]}

    In right triangle APB\!:\;PB \,=\,12-x
    . . \cos\theta \:=\:\frac{12-x}{15}\;\;{\color{blue}[2]}

    Equate [1] and [2]: . \frac{\sqrt{64-x^2}}{8} \;=\;\frac{12-x}{15}\quad\Rightarrow\quad 15\sqrt{64-x^2} \;=\;8(12-x)

    Square both sides: . 225(64-x^2) \;=\;64(144-24x + x^2)

    . . which simplifies to: . 289x^2 - 1536x - 5184 \;=\;0

    . . and has the positive root: . x \;=\;\frac{1536 + \sqrt{6,352,000}}{578} \;=\;7.657439446


    Then: . \sin\theta \;=\;\frac{x}{8} \;=\;\frac{7.657439446}{8} \;=\;0.957179931

    . . \theta \;=\;\sin^{-1}(0.957179931) \;=\;73.17235533^o \;\approx\;\boxed{73^o10'}

    Follow Math Help Forum on Facebook and Google+

  5. #5
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    I've got some difficulties, how can we know, by reading the text, what angle we are looking for ? And how can this physically be possible if D is not on RS ??
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Jan 2008
    Posts
    588
    Thanks
    87
    When I saw this question, I knew I saw it before. It was in my textbook last year. Here is the picture the book provided, and the answer at the back of the book is 16 degrees 50 minutes, not 16 degrees 15 minutes. I think Soroban had a correct answer, but the book is asking for the angle of the other side. So:

    180^o(straight angle) - 90^o (angle between base and length of cylinder) -  73^o10' (angle 'on the other side' Soroban found) = \theta

    \theta = 16^o 50'
    Attached Thumbnails Attached Thumbnails Trigonometry three dimensional question-untitled-1.jpg  
    Last edited by Gusbob; April 26th 2008 at 04:54 PM.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Jun 2006
    From
    NSW Australia
    Posts
    12

    Thanks for the responses

    Thanks everyone especially Soroban, you are genius. Two dimensional diagram is very helpful to understand. Thanks Gusbob for posting the diagram from the text book, I did not have scanner to scan the diagram.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Basis Dimensional analysis Question
    Posted in the Advanced Math Topics Forum
    Replies: 4
    Last Post: October 4th 2010, 04:10 AM
  2. Replies: 1
    Last Post: February 25th 2010, 01:15 AM
  3. Replies: 3
    Last Post: August 31st 2009, 11:48 AM
  4. Trigonometry Question
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: December 1st 2008, 12:07 PM
  5. trigonometry question.
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: December 19th 2007, 04:12 PM

Search Tags


/mathhelpforum @mathhelpforum