.Sam.
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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.Sam.
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dpaInc
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dpaInc
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.Sam.
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Actually you can simplify a little bit by
\[A(shaded)=\frac{1}{2}r^2(pi-\theta)-\frac{1}{2}(r^2)(\sin(pi-\theta))\]
\[A(shaded)=\frac{1}{2}(r^2)(pi-\theta-\sin(pi-\theta))\]
dpaInc
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magic happens in between...
dpaInc
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.Sam.
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POB=2AP
\[\frac{1}{2}r^2 \theta=2(\frac{1}{2}r^2(pi-\theta-\sin(pi-\theta))\]
\[\theta=2(pi-\theta-\sin(pi-\theta))\]
\[\theta=2pi-2\theta-2\sin(pi-\theta))\]
\[3\theta=2(pi-\sin(pi-\theta))\]
dpaInc
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yea, i missed that negative sin up there.
dpaInc
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nice problem @.Sam. .... thankx. :)
.Sam.
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I'll post another one but a bit mind boggling :D