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Abhisar

  • one year ago

Prove that length of the chord P1P2 is 2RCos \(\theta\)

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  1. Abhisar
    • one year ago
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    |dw:1437466437909:dw|

  2. welshfella
    • one year ago
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    |dw:1437471838416:dw|

  3. welshfella
    • one year ago
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    By the Cosine Rule r^2 =r^2 +p1p2 ^2 - 2r(p1p2)cos theta solve for p1p2

  4. welshfella
    • one year ago
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    p1p2 ^2 - 2r(p1p2)cos theta r^2 - r^2 = 0 p1p2(p1p2 - 2r cos theta) = 0 (p1p2 - 2r cos theta) = 0 or pip2 = 0 p1p2 = 2r cos theta

  5. welshfella
    • one year ago
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    * correction to first line:- p1p2 ^2 - 2r(p1p2)cos theta = r^2 - r^2 = 0

  6. Loser66
    • one year ago
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    |dw:1437481488578:dw|

  7. Loser66
    • one year ago
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    O is middle point of \(P_1P_2\), and \(P_1C= R\) Hence \(CO\perp P_1P_2\), that gives us \(cos (\theta)=\dfrac{P_1O}{R}\rightarrow 2P_1O=P_1P_2= 2Rcos(\theta)\)

  8. Abhisar
    • one year ago
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    Thank you very much @welshfella and @loser66

  9. Loser66
    • one year ago
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    @Astrophysics Make question here, please.

  10. Astrophysics
    • one year ago
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    Everything checks out, I had a similar method, but yours is quicker, nice job! :P

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