Results 1 to 6 of 6

Math Help - help please mechanics

  1. #1
    Newbie
    Joined
    Nov 2006
    Posts
    20

    Post help please mechanics

    a steel chimney is held in position by several guy wires. two such wires are AB and AC where B, C and the base of the chimney D, are on the same level in a straight line with B and C on the left of the chimney. if angle ABC=42 degrees, ACD=67degrees and BC=32M find the height of A above D..
    many thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123
    Quote Originally Posted by jim49990 View Post
    a steel chimney is held in position by several guy wires. two such wires are AB and AC where B, C and the base of the chimney D, are on the same level in a straight line with B and C on the left of the chimney. if angle ABC=42 degrees, ACD=67degrees and BC=32M find the height of A above D..
    many thanks
    Hello, Jim,

    1. Calculate the angle \angle(BAC)=180^\circ-42^\circ-67^\circ=71^\circ

    2. Calculate AB using Sine rule:
    \frac{AB}{32m}=\frac{\sin(67^\circ)}{\sin(71^\circ  )}
    AB\approx 31.15m

    3. Triangle BDA is a right triangle.
    \frac{AD}{AB}=\sin(42^\circ)\ \Longrightarrow\ AD\approx 20.84m

    EB
    Attached Thumbnails Attached Thumbnails help please mechanics-gei_an_schornst1.gif  
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by earboth View Post
    Hello, Jim,

    1. Calculate the angle \angle(BAC)=180^\circ-42^\circ-67^\circ=71^\circ

    2. Calculate AB using Sine rule:
    \frac{AB}{32m}=\frac{\sin(67^\circ)}{\sin(71^\circ  )}
    AB\approx 31.15m

    3. Triangle BDA is a right triangle.
    \frac{AD}{AB}=\sin(42^\circ)\ \Longrightarrow\ AD\approx 20.84m

    EB
    A and C are supposed to be on the same side of the chimney. My interpretation
    of the problem is shown in the attachment (however I still find the wording confusing).

    Anyway in this case you get:

    <br />
\tan(67)=AD/CD<br />

    and:

    <br />
\tan(42)=AD/(CD+32)<br />

    which gives:

    <br />
CD=AD/\tan(67)<br />
and CD=AD/\tan(42) -32

    Which gives the equation:

    <br />
AD/\tan(67)=AD/\tan(42) -32

    Which gives:

    <br />
AD=32/(\cot(42)-\cot(67))<br />


    RonL
    Attached Thumbnails Attached Thumbnails help please mechanics-gash.jpg  
    Last edited by CaptainBlack; November 19th 2006 at 01:14 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Nov 2006
    Posts
    20

    mant thanks

    thanks for that
    jim
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Nov 2006
    Posts
    20

    it also asks?

    it also asks to discuss an alternative approach and apply it..
    any ideas
    thanks
    jim
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by jim49990 View Post
    it also asks to discuss an alternative approach and apply it..
    any ideas
    thanks
    jim
    With a protractor to measure the angles construct a scale diagram
    and measure the height of the diagram.

    On my diagram the height is 46mm, where 1mm is equivalent to 1m, so
    this method for me gives an answer of 46m.

    This compares with an answer of ~=46.64m from the method I gave
    previously.

    A larger drawing (1cm to 1m lets say) should give a extra significant digit,
    and more care with the diagram might give some more accuracy (say
    to +/-0.025m).

    (a larger diagram would give more accuracy - also the angles could
    be constructed using the trig ratios rather than a protractor if need
    be).

    RonL
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Mechanics again
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: October 10th 2009, 10:10 AM
  2. Mechanics
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: October 9th 2009, 04:20 AM
  3. Maths Mechanics SUVAT mechanics
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 27th 2009, 04:55 PM
  4. Maths Mechanics SUVAT mechanics
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: April 27th 2009, 03:44 PM
  5. Mechanics one
    Posted in the Advanced Applied Math Forum
    Replies: 2
    Last Post: March 15th 2009, 09:44 AM

Search Tags


/mathhelpforum @mathhelpforum