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circle of radius r.

I think you will benefit from this equation: A= 1/2 L^2 sinθ

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- Thread starter vip89
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In summary: Presumably, this person knows more about calculus than you do. If you want to get serious about calculus, you should consult with a professional.

- #1

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circle of radius r.

I think you will benefit from this equation: A= 1/2 L^2 sinθ

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- #3

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I know that,but how I can make it with one variable??

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Half of the isosceles triangle is a right triangle with hypotenuse L and one side [itex]H= L cos(\theta/2)[/itex], the altitude of the triangle, and the third side [itex]L sin(\theta/2)[/itex]. If you draw a line from the center of the circle to point at which the line L is tangent to the circle, that also gives a right triangle (a radius of a circle is always perpendicular to a tangent) similar to the first right triangle. The hypotenuse of this smaller right triangle is H-r and the "opposite side" has length r. That is

[tex]\frac{r}{H-r}= \frac{L sin(\theta/2)}{L}[/tex]

That will allow you to write the formula in terms of the variable [itex]\theta[/itex] and r, which is a constant.

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vip89 said:That is my try

You do not really expect anybody to read this do you? I suppose you would not hand in something like this to your professor or TA or teacher or whoever (at least I strongly hope so) so why don't you bother writing it down in a tidy way if you want us to check your work?

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but doctor said that he want a calculus??

I didnt know how to solve by calculus?

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- #9

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1. Surely the "school method" is not to write at random angles all across the paper!vip89 said:

but doctor said that he want a calculus??

I didnt know how to solve by calculus?

2. Who is "the doctor"?

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