## .Sam. 3 years ago Proving easy one, Area of sector POB is equal to twich the area of the shaded region, show that $3\theta=2(pi-sin( pi-\theta))$

1. .Sam.

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2. dpaInc

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3. dpaInc

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4. .Sam.

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

magic happens in between...

6. dpaInc

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7. .Sam.

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

yea, i missed that negative sin up there.

9. dpaInc

nice problem @.Sam. .... thankx. :)

10. .Sam.

I'll post another one but a bit mind boggling :D