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Math Help - Can't figure these 2 out.......angle and distance

  1. #1
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    Can't figure these 2 out.......angle and distance

    See the attached form. Problem 3 and 5. I have tried several times and can't seem to figure out how to get the angle in 3 and the distance in 5.

    I got 5 with an answer of 5.0176 inches and it's 5.024 inches.
    I STILL can't get 3 with all angles I can find and lengths.!!!!!


    Thanks for any help given.
    Joanne
    Attached Thumbnails Attached Thumbnails Can't figure these 2 out.......angle and distance-question-3-5.jpg  
    Last edited by bradycat; February 26th 2011 at 08:27 AM.
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  2. #2
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    For 3, you have R\sin(90-y)=R\sin(42^\circ50')+1.480, from where y can be found.

    Can't figure these 2 out.......angle and distance-draft.png
    For 5, \tan(57/2)=\tan\angle OAC=OC/AC, from where AC can be found. (See the attached image above.) Also, \tan 57=DB/AB, from where AB can be found. So, we know the horizontal location of A (3.400 - AB) and therefore the location of C.
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  3. #3
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    still stuck on the #3, Not getting what your saying. Can you explain a bit more on calculating that please, thanks.
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  4. #4
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    Are you not sure about the equation I wrote or how to solve it? Let x be the angle between y and 42^\circ50'. Then R\sin(x+42^\circ50') is the vertical coordinate of the center of the top circle. It is equal to 1.480 plus the vertical coordinate of the center of the rightmost circle. The latter is R\sin42^\circ50'. In all this, R = 4.646.

    To find y, divide both sides by R and take arcsine of both sides. What I get, though, is y = 3^\circ13', not 3^\circ12', as the answer seems to say.
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  5. #5
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    Sorry I am having trouble solving it. I am not getting anything close. Can you show the steps please, thank you for your help. Some reason I am not seeing thing.
    Jo
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  6. #6
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    You still have not said whether you understand why R\sin(90-y)=R\sin(42^\circ50')+1.480 is true. Assuming you do, divide both sides by R = 4.646 to get \sin(90-y)=\sin(42^\circ50')+1.480/4.646. Taking \sin^{-1} of both sides gets 90-y=\sin^{-1}(\sin(42^\circ50')+1.480/4.646). Since 42^\circ50'=(42+5/6)^\circ, evaluating the right-hand side gives 86.780^\circ, i.e., y=3.220^\circ. Finally, 0.220\cdot 60\approx 13, so 3.22^\circ=3^\circ 13'.

    If something else is not clear, could you specify precisely what?
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