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

[luigihs]

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?

 

[rollcast]

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

\ c=pi*d

and then multiplied that answer by,

\frac{60}{360}

 

[luigihs]

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

\ c=pi*d

and then multiplied that answer by,

\frac{60}{360}

Ok but my answer is right?

 

[rollcast]

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?

 

[luigihs]

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 ??

 

[rollcast]

A = 60 / 360 x 2 ∏ 3 ??

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

So for the sector area your equation should be,

Area of Sector = \pi r ^ {2} * \frac{Angle of Sector°}{360°}

 

[luigihs]

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

So for the sector area your equation should be,

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

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

 

[rollcast]

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

Perfect.

 

[luigihs]

Perfect.

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

 

[rollcast]

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)

x - \frac{x^ {3}}{3!} + \frac{x^ {5}}{5!} -\frac{x^ {7}}{7!} + ...

For cos(x)

1 - \frac{x^ {2}}{2!} + \frac{x^ {4}}{4!} -\frac{x^ {6}}{6!} +...

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

 

[rollcast]

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

 

[Mark44]

I found this on another website.

For sin(x)

x - \frac{x^ {3}}{3!} + \frac{x^ {5}}{5!} -\frac{x^ {7}}{7!} + ...

For cos(x)

1 - \frac{x^ {2}}{2!} + \frac{x^ {4}}{4!} -\frac{x^ {6}}{6!} +...

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

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

 

[rollcast]

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