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Math Help - Vectors and Oblique Triangles

  1. #1
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    Vectors and Oblique Triangles

    can someone help me on this? thanks alot!!

    add the given vectors by using the trigonometric functions and the pythagorean theorem

    A naval cruiser on maneuvers travels 54.0km at 18.7 degree west of north, then turns and travels 64.5km at 15.6 degree south of east, and finally turns to travel 72.4km at 38.1 degree east of south. Find its displacement from its original position.
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  2. #2
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    Hello, Daniel!

    Add the given vectors by using the trigonometric functions and the pythagorean theorem
    You're kidding, right?
    A naval cruiser on maneuvers travels 54.0km at 18.7░ west of north,
    then turns and travels 64.5km at 15.6░ south of east,
    and finally turns to travel 72.4km at 38.1░ east of south.
    Find its displacement from its original position.

    Let's work this out in baby-steps.

    The cruiser went from A to B.
    It moved: 54Ěsin18.7░ km west and 54Ěcos18.7░ km north.
    Code:
              54Ěsin18.7░
        B * - - - - - - - :
            *             :
              *           :
             54 *         : 54Ěcos18.7░
                  * 18.7░ :
                    *     :
                      *   :
                        * :
                          * A

    Then it went from B to C.
    It moved: 64.5Ěcos15.6░ km east and 64.5Ěsin15.6░ km south.
    Code:
                64.5.cos15.6░
        B * - - - - - - - - - +
              *  15.6░        :
                  *           : 64.5.sin15.6░
                 64.5 *       :
                          *   :
                              * C

    Then it went from C to D.
    It moved: 72.4Ěsin38.1░ km east and 72.4Ěcos38.1░ km south.
    Code:
                C *
                  :  *
                  :38.1░*
                  :        * 72.4
    72.4Ěcos38.1░ :           *
                  :              *
                  + - - - - - - - - * D
                      72.4Ěsin38.1░


    Can you determine the location of the cruiser now?

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  3. #3
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    Since we are finding a displacement, all you are really doing is adding up all of the x components and all of the y components. So first, we need to find all of the degree measures from the x-axis, since it is the easiest (108.7, 341.3, and 308.1 respectively). After that, take the cosine of each angle multiplied by the length of each vector (cos(108.7)*54, cos(341.3)*64.5, and cos(308.1)*72.4) and add them up. You should get approximately 88.46. Now do the same thing, except taking the sine of each angle times the length, which is approximately -26.50. These are the two legs of your right triangle. Now use the pythagoream theorem to find your missing length. This is the length of the displacement (92.34km). To find the angle of the displacement, take the inverse tangent of your legs (tan^-1(88.46/-26.50)), and take the absolute value, which is 73.32 degrees east of south. All done!
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  4. #4
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    Quote Originally Posted by Soroban View Post
    Hello, Daniel!



    Let's work this out in baby-steps.

    The cruiser went from A to B.
    It moved: 54Ěsin18.7░ km west and 54Ěcos18.7░ km north.
    Code:
              54Ěsin18.7░
        B * - - - - - - - :
            *             :
              *           :
             54 *         : 54Ěcos18.7░
                  * 18.7░ :
                    *     :
                      *   :
                        * :
                          * A

    Then it went from B to C.
    It moved: 64.5Ěcos15.6░ km east and 64.5Ěsin15.6░ km south.
    Code:
                64.5.cos15.6░
        B * - - - - - - - - - +
              *  15.6░        :
                  *           : 64.5.sin15.6░
                 64.5 *       :
                          *   :
                              * C

    Then it went from C to D.
    It moved: 72.4Ěsin38.1░ km east and 72.4Ěcos38.1░ km south.
    Code:
                C *
                  :  *
                  :38.1░*
                  :        * 72.4
    72.4Ěcos38.1░ :           *
                  :              *
                  + - - - - - - - - * D
                      72.4Ěsin38.1░


    Can you determine the location of the cruiser now?


    If i use soroban method i will get
    <br />
rx = ax + bx + cx, <br />
ry = ay + bY + cy<br />
rx = 51.2 + 62.1 + 57<br />
rx = 170.3<br />
ry = 17.3 + 17.4 + 44.7<br />
ry = 79.4<br />
r = sqrt (170.3)^2 + (79.4)^2<br />
r = 187.9<br /> <br />
tan-1 (79.4/170.3) = 25 degree<br /> <br />
am i doing it the right way?<br /> <br />
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  5. #5
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    Quote Originally Posted by dnlstffrd View Post
    Since we are finding a displacement, all you are really doing is adding up all of the x components and all of the y components. So first, we need to find all of the degree measures from the x-axis, since it is the easiest (108.7, 341.3, and 308.1 respectively). After that, take the cosine of each angle multiplied by the length of each vector (cos(108.7)*54, cos(341.3)*64.5, and cos(308.1)*72.4) and add them up. You should get approximately 88.46. Now do the same thing, except taking the sine of each angle times the length, which is approximately -26.50. These are the two legs of your right triangle. Now use the pythagoream theorem to find your missing length. This is the length of the displacement (92.34km). To find the angle of the displacement, take the inverse tangent of your legs (tan^-1(88.46/-26.50)), and take the absolute value, which is 73.32 degrees east of south. All done!
    anyway how do you got this??? ----> (108.7, 341.3, and 308.1 respectively).
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  6. #6
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    draw it out. 18.7 degrees west of north is going to be in the second quadrant, so add 90 degrees. 15.6 degrees south of east is going to be just less than 360 degrees, so it is just 360-15.6. 38.1 degrees is also going ot be in the fourth quadrant, so take 270+38.1. That is how I got those respective values.
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  7. #7
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    Quote Originally Posted by dnlstffrd View Post
    draw it out. 18.7 degrees west of north is going to be in the second quadrant, so add 90 degrees. 15.6 degrees south of east is going to be just less than 360 degrees, so it is just 360-15.6. 38.1 degrees is also going ot be in the fourth quadrant, so take 270+38.1. That is how I got those respective values.
    so did soraban do it wrong??
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  8. #8
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    This is the method I have been tought by 2 teachers and has worked very well so far. It appears to me that he used the sin of 2 of the angles to find the horizontal distance, when I have been taught to use cos, since if you look at a unit circle, cosine is related to x, not y.
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  9. #9
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by dnlstffrd View Post
    Twhen I have been taught to use cos, since if you look at a unit circle, cosine is related to x, not y.
    I haven't explored this problem in any detail so I don't know if Soroban did it right or not, but I want to warn you NOT to associate cos with the x direction and sin with the y direction. The angles you are given may or may not correspond this way. I recommend to always sketch the triangles and use the sin and cos definitions to determine which of x, y is associated with sin, cos.

    Take it from a Physics teacher who is always seeing students get mixed up about this.

    -Dan
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  10. #10
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    but most of the time you can make it in a way that cos does relate to x. and you must actually be a good physics teacher, mine sucked on this topic. He only knew the method of drawing it out. When it came to calculations, I learned everything from my Precalc teacher.
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  11. #11
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by dnlstffrd View Post
    but most of the time you can make it in a way that cos does relate to x. and you must actually be a good physics teacher, mine sucked on this topic. He only knew the method of drawing it out. When it came to calculations, I learned everything from my Precalc teacher.
    I humbly hope I can be considered such. Yes, most of the time you are right. It's particularly when you get to the topic of forces and torques that you really need to be careful about where the angles are. With vector addition you can usually get away with it.

    However in this case:
    54.0km at 18.7 degree west of north
    and
    72.4km at 38.1 degree east of south
    will give you sin associated with x and cos associated with y if you use the typical coordinate system where E is +x and N is +y. Soroban's diagrams were correct there, so I'm assuming his use of sin and cos were as well.

    -Dan
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