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Thread: 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; Feb 26th 2011 at 07:27 AM.
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  2. #2
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    For 3, you have $\displaystyle R\sin(90-y)=R\sin(42^\circ50')+1.480$, from where y can be found.

    Can't figure these 2 out.......angle and distance-draft.png
    For 5, $\displaystyle \tan(57/2)=\tan\angle OAC=OC/AC$, from where AC can be found. (See the attached image above.) Also, $\displaystyle \tan 57=DB/AB$, 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 $\displaystyle 42^\circ50'$. Then $\displaystyle R\sin(x+42^\circ50')$ 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 $\displaystyle R\sin42^\circ50'$. 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 $\displaystyle y = 3^\circ13'$, not $\displaystyle 3^\circ12'$, 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 $\displaystyle R\sin(90-y)=R\sin(42^\circ50')+1.480$ is true. Assuming you do, divide both sides by R = 4.646 to get $\displaystyle \sin(90-y)=\sin(42^\circ50')+1.480/4.646$. Taking $\displaystyle \sin^{-1}$ of both sides gets $\displaystyle 90-y=\sin^{-1}(\sin(42^\circ50')+1.480/4.646)$. Since $\displaystyle 42^\circ50'=(42+5/6)^\circ$, evaluating the right-hand side gives $\displaystyle 86.780^\circ$, i.e., $\displaystyle y=3.220^\circ$. Finally, $\displaystyle 0.220\cdot 60\approx 13$, so $\displaystyle 3.22^\circ=3^\circ 13'$.

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