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Thread: Tricky triogonometri...

  1. #1
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    Tricky triogonometri...

    How can I find angle v in this figure?

    Info:

    AO=OB=OP=OC=6370 km
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  2. #2
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    Tricky with the figure missing?
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  3. #3
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    continues....

    PS=20240 km
    CP = 2780 km

    Thanks for all answers
    Attached Thumbnails Attached Thumbnails Tricky triogonometri...-gash.jpg  
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  4. #4
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    Quote Originally Posted by Neffets...:P
    PS=20240 km
    CP = 2780 km

    Thanks for all answers
    The key is that $\displaystyle \angle POC=(arcCP)/OC=2780/6370$ radian.

    The rest is just fiddly trig.
    Last edited by CaptainBlack; Aug 26th 2006 at 08:49 AM.
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  5. #5
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    thanks

    thanks mate! I think i'll manage it all now...



    You guys are GREAT!
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  6. #6
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    need more help...

    i have tried a lot, but i have not been able to find the answer!

    CP is a part of a circle...
    the angle is on a tangent.

    Can you guys show me how you find the answer?
    I know I seem a bit stupid, but...

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  7. #7
    Grand Panjandrum
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    Quote Originally Posted by Neffets...:P
    i have tried a lot, but i have not been able to find the answer!

    CP is a part of a circle...
    the angle is on a tangent.

    Can you guys show me how you find the answer?
    I know I seem a bit stupid, but...

    Drop a perpendicular from $\displaystyle C$ onto $\displaystyle OS$, call the new point at the foot of the perp. $\displaystyle Q$.

    Now look at $\displaystyle \triangle OCQ$ you know $\displaystyle \angle QOC$, and $\displaystyle OC$, so you can find $\displaystyle OQ$ and $\displaystyle CQ$ using the $\displaystyle \sin$ and $\displaystyle \cos$ of $\displaystyle \angle QOC$ and $\displaystyle OC$.

    Now switch attention to $\displaystyle \triangle CQS$. You know $\displaystyle SO$ and $\displaystyle OQ$, so you know $\displaystyle QS$. You already know $\displaystyle CQ$, so Pythagoras's theorem will now finish the problem.

    RonL
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  8. #8
    Grand Panjandrum
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    Quote Originally Posted by CaptainBlack
    Drop a perpendicular from $\displaystyle C$ onto $\displaystyle OS$, call the new point at the foot of the perp. $\displaystyle Q$.

    Now look at $\displaystyle \triangle OCQ$ you know $\displaystyle \angle QOC$, and $\displaystyle OC$, so you can find $\displaystyle OQ$ and $\displaystyle CQ$ using the $\displaystyle \sin$ and $\displaystyle \cos$ of $\displaystyle \angle QOC$ and $\displaystyle OC$.

    Now switch attention to $\displaystyle \triangle CQS$. You know $\displaystyle SO$ and $\displaystyle OQ$, so you know $\displaystyle QS$. You already know $\displaystyle CQ$, so Pythagoras's theorem will now finish the problem.

    RonL
    I forgot:

    Extent the tangent at $\displaystyle C$ to meet $\displaystyle OS$ at $\displaystyle T$, then $\displaystyle \triangle OCQ$ is similar to $\displaystyle \triangle CQT$. This will allow you to find $\displaystyle \angle CPS$, and you already have enough information to fins $\displaystyle \angle PSC$ which will allow you to find angle at $\displaystyle v$

    RonL
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  9. #9
    Grand Panjandrum
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    The attached scan shows all the relevant dimensions and angles filled
    in on the diagram.

    RonL
    Attached Thumbnails Attached Thumbnails Tricky triogonometri...-gash.jpg  
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  10. #10
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    Talking Thanks Captain!

    This forum is great! Now I can finally get help when I'm in a tricky situation...
    Thanks a lot Captain Black!!!! :-D :-D

    I couldn't find the answer with your first explanation, so I'm glad you did remember the rest...


    Neffets...:P
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