Determine the lenght of arc and the area of the sector subtended by an

In summary: Also I forgot to say that you need to be using radian values for those 2 series to work.Ok, thank you!In summary, Determine the length of arc and the area of the sector subtended by an angle of 60° in circle of radius 3 m with the help of the Maclaurin series for sin and cos.
  • #1
luigihs
86
0
Determine the length of arc and the area of the sector subtended by an angle of 60° in circle of radius 3 m

Ok First change 60to radian measure. 60 x pi / 180° = 2pi / 6

Then.. I used the formula s = rθ 3(2pi/6) = 6pi/6 = pi <--- Is this right??

And how can I find the area which formula can I use?
 
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  • #2


You could just have kept the angle in degrees and worked out the arc length by using,

[itex]\ c=pi*d[/itex]

and then multiplied that answer by,

[itex]\frac{60}{360}[/itex]
 
  • #3


rollcast said:
You could just have kept the angle in degrees and worked out the arc length by using,

[itex]\ c=pi*d[/itex]

and then multiplied that answer by,

[itex]\frac{60}{360}[/itex]

Ok but my answer is right?
 
  • #4


Yes your answer is correct.

That might give you a clue how to find the area of the sector. Think of the fraction of the total circles area you are looking for?
 
  • #5


rollcast said:
Yes your answer is correct.

That might give you a clue how to find the area of the sector. Think of the fraction of the total circles area you are looking for?

A = 60 / 360 x 2 ∏ 3 ??
 
  • #6


luigihs said:
A = 60 / 360 x 2 ∏ 3 ??

Nearly. You're along the right lines but your calculation for the total area is wrong, [itex]\ a = pi * r^{2}[/itex]

So for the sector area your equation should be,

[itex] Area of Sector = \pi r ^ {2} * \frac{Angle of Sector°}{360°}[/itex]
 
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  • #7


rollcast said:
Nearly. You're along the right lines but your calculation for the total area is wrong, [itex]\ a = pi * r^{2}[/itex]

So for the sector area your equation should be,

[itex]\ Area of Sector = ( pi * r ^ {2} )* (\frac{Angle of sector°}{360°})[/itex]

Ok so A = 9 pi x 60 / 360 = 540 pi / 360 = 3pi / 2
 
  • #8


luigihs said:
Ok so A = 9 pi x 60 / 360 = 540 pi / 360 = 3pi / 2

Perfect.
 
  • #9


rollcast said:
Perfect.

Yay! Hey do you know how to determine cos or sin without calculator?? like cos 75° any hint?
 
  • #10


luigihs said:
Yay! Hey do you know how to determine cos or sin without calculator?? like cos 75° any hint?

I found this on another website.

For sin(x)

[itex] x - \frac{x^ {3}}{3!} + \frac{x^ {5}}{5!} -\frac{x^ {7}}{7!} + ...[/itex]

For cos(x)

[itex] 1 - \frac{x^ {2}}{2!} + \frac{x^ {4}}{4!} -\frac{x^ {6}}{6!} +...[/itex]

From a bit of quick testing here these series seem to converge to the right value fairly quickly.

Not sure if that really helps you much

AL
 
  • #11


Also I forgot to say that you need to be using radian values for those 2 series to work.

edit.

I just realized you would still need a basic calculator to do that.

Without a calculator of any sort your options are really either getting a trig table sheet or learning some of the common ones like 0, 30, 45, 60, 90 etc
 
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  • #12


rollcast said:
I found this on another website.

For sin(x)

[itex] x - \frac{x^ {3}}{3!} + \frac{x^ {5}}{5!} -\frac{x^ {7}}{7!} + ...[/itex]

For cos(x)

[itex] 1 - \frac{x^ {2}}{2!} + \frac{x^ {4}}{4!} -\frac{x^ {6}}{6!} +...[/itex]
These are the Maclaurin series for the sine and cosine functions. Maclaurin series are special cases of Taylor series.
rollcast said:
From a bit of quick testing here these series seem to converge to the right value fairly quickly.

Not sure if that really helps you much

AL
 
  • #13


Mark44 said:
These are the Maclaurin series for the sine and cosine functions. Maclaurin series are special cases of Taylor series.

Thanks Mark, I haven't really studied series in school in much detail yet so thanks for telling me what those where.

Thanks
AL
 

1. What is the formula for determining the length of an arc?

The formula for determining the length of an arc is L = rθ, where L is the length of the arc, r is the radius of the circle, and θ is the central angle in radians.

2. How do you find the area of a sector?

The formula for finding the area of a sector is A = ½r²θ, where A is the area of the sector, r is the radius of the circle, and θ is the central angle in radians.

3. Can you use degrees instead of radians to find the length of an arc and area of a sector?

Yes, you can convert degrees to radians by using the formula θ = π/180 * α, where θ is the angle in radians and α is the angle in degrees. Then, you can use the formula L = rθ and A = ½r²θ to find the length of the arc and area of the sector.

4. How do you determine the length of an arc or area of a sector for a fraction of a circle?

To determine the length of an arc or area of a sector for a fraction of a circle, you need to modify the formulas to include the fraction of the circle. For example, if you want to find the length of an arc that is ⅓ of a circle, you would use the formula L = (r * 2π * ⅓) / 2π = r / 3.

5. Can the length of an arc or area of a sector be negative?

No, the length of an arc and area of a sector cannot be negative. These values represent physical measurements and cannot have negative values. If you get a negative value when using the formulas, it may indicate an error in your calculations.

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