# Maths Aid!

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#21

(Original post by

If the sector is in a circle of radius r and has an angle t (in radians), then:

Area = ( tr^2 )/2

Perimeter = 2r + rt = r(t+2)

So a square with the same perimeter has sides of length r(t+2)/4; therefore, the area of this square is ( r(t+2)/4 )^2.

We require the area of the sector to be equal to the area of the square, so:

( tr^2 )/2 = ( r(t+2)/4 )^2

( tr^2 )/2 = ( r^2(t+2)^2 ) /16

8tr^2 = r^2(t+2)^2

The r^2 terms cancel, so:

8t = (t+2)^2

8t = t^2 + 4t + 4

t^2 -4t +4 = 0

(t-2)^2 = 0

t = 2

**mikesgt2**)If the sector is in a circle of radius r and has an angle t (in radians), then:

Area = ( tr^2 )/2

Perimeter = 2r + rt = r(t+2)

So a square with the same perimeter has sides of length r(t+2)/4; therefore, the area of this square is ( r(t+2)/4 )^2.

We require the area of the sector to be equal to the area of the square, so:

( tr^2 )/2 = ( r(t+2)/4 )^2

( tr^2 )/2 = ( r^2(t+2)^2 ) /16

8tr^2 = r^2(t+2)^2

The r^2 terms cancel, so:

8t = (t+2)^2

8t = t^2 + 4t + 4

t^2 -4t +4 = 0

(t-2)^2 = 0

t = 2

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#22

(Original post by

They clean up lots of equations. If you want to use degrees you very often end up with a 360/2pi factor.

Also, when you need to raise a number to a complex exponent you probably want to use radians when expressing the complex numbers in polar form. This gives rise the following beautiful equation:

e^(i * pi) + 1 = 0

Also, do you take this task (Its a physics one, but still)

At what height do you find geostationary satelites?

**Jonatan**)They clean up lots of equations. If you want to use degrees you very often end up with a 360/2pi factor.

Also, when you need to raise a number to a complex exponent you probably want to use radians when expressing the complex numbers in polar form. This gives rise the following beautiful equation:

e^(i * pi) + 1 = 0

Also, do you take this task (Its a physics one, but still)

At what height do you find geostationary satelites?

Consider the satellite to have mass m, height h and velocity v and the earth to have mass M, radius r and time period T

force due to circular motion=force due to gravitational attraction

(mv^2)/(r+h)=GMm/(r+h)^2

v^2=GM/(r+h)

v=(GM/(r+h))^0.5

The time period can be expressed as

T=2π(r+h)/v

v=2π(r+h)/T

Put both equations equal to eachother and rearrange.

(GM/(r+h))^0.5=2π(r+h)/T

GM/(r+h)=4π^2(r+h)^2/T^2

(r+h)^3=GMT^2/4π^2

r=(GMT^2/4π^2)^(1/3)-r

Substitute in T = 24*60*60= 86400s, G = 6.67*10-11 N m2/kg2, M = 5.97*10^24kg and R = 6.37*10^6m, and then h=3.59*10^7m

Oh, and e^(i*pi)+1=0, derived from Taylor expansions - again differentiating trig functions. I will admit they tidy things up, though.

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#23

(Original post by

My ridiculously complicated method didn't because I wasn't integrating trig functions, that's the only time I can think of when radians are useful.

**sazzles**)My ridiculously complicated method didn't because I wasn't integrating trig functions, that's the only time I can think of when radians are useful.

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#24

(Original post by

Radians are essential as, unlike degrees, they represent angles and lengths therefore equations such as y = xcosx, y = x - tanx etc can only be studied using radians. You cannot have a length of 360 degrees but you can have a length of 2pi. So obviously in calculus where you say as x->0, sinx -> x you can only use radians because you have changed x from an angle to an approximate length

**It'sPhil...**)Radians are essential as, unlike degrees, they represent angles and lengths therefore equations such as y = xcosx, y = x - tanx etc can only be studied using radians. You cannot have a length of 360 degrees but you can have a length of 2pi. So obviously in calculus where you say as x->0, sinx -> x you can only use radians because you have changed x from an angle to an approximate length

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#25

I think that this problem I did in class today is quite neat, it's an extension of the sine rule:

Say that ABC is a triangle and P is an point inside ABC. Let a = <BPC - <BAC, b = <CPA - <CBA, c = <APB - <ACB. ( < means angle here).

Show that (PA.sin<BAC)/sin(a) = (PB.sin<CBA)/sin(b) = (PC.sin<ACB)/sin(c).

Say that ABC is a triangle and P is an point inside ABC. Let a = <BPC - <BAC, b = <CPA - <CBA, c = <APB - <ACB. ( < means angle here).

Show that (PA.sin<BAC)/sin(a) = (PB.sin<CBA)/sin(b) = (PC.sin<ACB)/sin(c).

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yeah nice,

see told u it was radians sazzles.

i got a good problem here take a look at this puzzle (it was easier to draw than type out) and you use common denominator to work out the second line, if u didnt understand

see told u it was radians sazzles.

i got a good problem here take a look at this puzzle (it was easier to draw than type out) and you use common denominator to work out the second line, if u didnt understand

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#27

(Original post by

I think that this problem I did in class today is quite neat, it's an extension of the sine rule:

Say that ABC is a triangle and P is an point inside ABC. Let a = <BPC - <BAC, b = <CPA - <CBA, c = <APB - <ACB. ( < means angle here).

Show that (PA.sin<BAC)/sin(a) = (PB.sin<CBA)/sin(b) = (PC.sin<ACB)/sin(c).

**theone**)I think that this problem I did in class today is quite neat, it's an extension of the sine rule:

Say that ABC is a triangle and P is an point inside ABC. Let a = <BPC - <BAC, b = <CPA - <CBA, c = <APB - <ACB. ( < means angle here).

Show that (PA.sin<BAC)/sin(a) = (PB.sin<CBA)/sin(b) = (PC.sin<ACB)/sin(c).

On the subject of triangles and formulae, you just reminded me of Heron's formula for the area of a triangle, A with sides of length a, b and c.

A=(s*(s-a)*(s-b)*(s-c))^(1/2)

where s=(a+b+c)/2

How random!?

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#28

please please please help me on this question, the exam is 2moro!

Find the value's of x between 0 and 180 satisfying the equation

tan3x=1

i know you tan-1 both sides to give 45 and divide by 3 because of the 3 before the x, but how do you find the other values? thanks!

Find the value's of x between 0 and 180 satisfying the equation

tan3x=1

i know you tan-1 both sides to give 45 and divide by 3 because of the 3 before the x, but how do you find the other values? thanks!

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#29

(Original post by

please please please help me on this question, the exam is 2moro!

Find the value's of x between 0 and 180 satisfying the equation

tan3x=1

i know you tan-1 both sides to give 45 and divide by 3 because of the 3 before the x, but how do you find the other values? thanks!

**greenie787**)please please please help me on this question, the exam is 2moro!

Find the value's of x between 0 and 180 satisfying the equation

tan3x=1

i know you tan-1 both sides to give 45 and divide by 3 because of the 3 before the x, but how do you find the other values? thanks!

Look at what other quadriles tan is positive (1st and 3th). This means that;

3x=45, 180+45, 360+45, 360+180+45, 2*360+45, 3*360+180+45 etc

3x=45, 225, 405, 585, 1035 etc

x=15, 75, 135, 195, 255, 345

Hope this is useful.

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#30

(Original post by

yeah nice,

see told u it was radians sazzles.

i got a good problem here take a look at this puzzle (it was easier to draw than type out) and you use common denominator to work out the second line, if u didnt understand

**LimpNoodle**)yeah nice,

see told u it was radians sazzles.

i got a good problem here take a look at this puzzle (it was easier to draw than type out) and you use common denominator to work out the second line, if u didnt understand

Graph the function f(x)=x/(x-1) + (x-l)/x -2

f(x)=2 is an asymptote.

I liked that one it was funny.

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#31

(Original post by

You done the first step, work out the acute angle (45)

Look at what other quadriles tan is positive (1st and 4th). This means that;

3x=45, 360-45, 360+45, 2*360-45, 2*360+45, 3*360-45 etc

3x=45, 315, 405, 675, 1035 etc

x=15, 105, 135, 225, 255, 345

Hope this is useful.

**sazzles**)You done the first step, work out the acute angle (45)

Look at what other quadriles tan is positive (1st and 4th). This means that;

3x=45, 360-45, 360+45, 2*360-45, 2*360+45, 3*360-45 etc

3x=45, 315, 405, 675, 1035 etc

x=15, 105, 135, 225, 255, 345

Hope this is useful.

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#32

(Original post by

thank you. with the sin and cos angles i draw the graphs, but with tan its a lot harder

**greenie787**)thank you. with the sin and cos angles i draw the graphs, but with tan its a lot harder

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#33

(Original post by

That's really interesting, I'd never thought of doing it that way. Only thing is, how does it show that this is the maximum area a sector could take?

**sazzles**)That's really interesting, I'd never thought of doing it that way. Only thing is, how does it show that this is the maximum area a sector could take?

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#34

(Original post by

Well, when you equate the area of the sector with the area of the square you find that there is only one angle for which the areas are equal. Since there is only one, it must be maximum.

**mikesgt2**)Well, when you equate the area of the sector with the area of the square you find that there is only one angle for which the areas are equal. Since there is only one, it must be maximum.

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#35

(Original post by

No problem. I agree with you about tan graphs, horrible things.

**sazzles**)No problem. I agree with you about tan graphs, horrible things.

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#36

(Original post by

Can you explain it again? I'm getting confused It seems as though you are using a circular argument.

**sazzles**)Can you explain it again? I'm getting confused It seems as though you are using a circular argument.

All I am saying is that you do not end up with an equation which you have to maximise using calculus. You simply do the maths and come up with the answer of 2 radians. The way I did the question seemed to show that there was only one angle for which the area of the sector was equal to the area of the square, so there is not an issue of maximum.

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#37

(Original post by

Why is it circular?

All I am saying is that you do not end up with an equation which you have to maximise using calculus. You simply do the maths and come up with the answer of 2 radians. The way I did the question seemed to show that there was only one angle for which the area of the sector was equal to the area of the square, so there is not an issue of maximum.

**mikesgt2**)Why is it circular?

All I am saying is that you do not end up with an equation which you have to maximise using calculus. You simply do the maths and come up with the answer of 2 radians. The way I did the question seemed to show that there was only one angle for which the area of the sector was equal to the area of the square, so there is not an issue of maximum.

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#38

Ok here's a pretty easy but quite fun problem - find digits a,b,c,d,e such that 4(abcde) = edcba

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hmmmmmm sounds interesting, but i have no idea what, must take a while unless there is a shortcut you can take, but you got me stumped

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