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.Sam.

  • 4 years ago

Proving easy one, Area of sector POB is equal to twich the area of the shaded region, show that \[3\theta=2(pi-sin( pi-\theta))\]

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  1. .Sam.
    • 4 years ago
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    |dw:1334895566070:dw|

  2. dpaInc
    • 4 years ago
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    |dw:1334897255916:dw|

  3. dpaInc
    • 4 years ago
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    |dw:1334897397459:dw|

  4. .Sam.
    • 4 years ago
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    Actually you can simplify a little bit by \[A(shaded)=\frac{1}{2}r^2(pi-\theta)-\frac{1}{2}(r^2)(\sin(pi-\theta))\] \[A(shaded)=\frac{1}{2}(r^2)(pi-\theta-\sin(pi-\theta))\]

  5. dpaInc
    • 4 years ago
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    magic happens in between...

  6. dpaInc
    • 4 years ago
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    |dw:1334897856911:dw|

  7. .Sam.
    • 4 years ago
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    POB=2AP \[\frac{1}{2}r^2 \theta=2(\frac{1}{2}r^2(pi-\theta-\sin(pi-\theta))\] \[\theta=2(pi-\theta-\sin(pi-\theta))\] \[\theta=2pi-2\theta-2\sin(pi-\theta))\] \[3\theta=2(pi-\sin(pi-\theta))\]

  8. dpaInc
    • 4 years ago
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    yea, i missed that negative sin up there.

  9. dpaInc
    • 4 years ago
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    nice problem @.Sam. .... thankx. :)

  10. .Sam.
    • 4 years ago
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    I'll post another one but a bit mind boggling :D

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