# Maths Aid!

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16 years ago
#21
(Original post by 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
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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16 years ago
#22
(Original post by 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?
A geostationary satellite goes around the earth with an angular velocity equal to that of earth.
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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16 years ago
#23
(Original post by 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.
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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16 years ago
#24
(Original post by 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
I see you point about the trig function, I was wrong;I don't really like radians (I always try to work things out in radians when my calculator is in degrees mode and vice versa). What you're saying about the approximations is the same my previous post-I only use radians for calculus.
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16 years ago
#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).
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#26
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
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16 years ago
#27
(Original post by 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).
That's really interesting.
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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16 years ago
#28

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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16 years ago
#29
(Original post by greenie787)

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!
You done the first step, work out the acute angle (45)
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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16 years ago
#30
(Original post by 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
lol.
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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16 years ago
#31
(Original post by 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.
thank you. with the sin and cos angles i draw the graphs, but with tan its a lot harder
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16 years ago
#32
(Original post by greenie787)
thank you. with the sin and cos angles i draw the graphs, but with tan its a lot harder
No problem. I agree with you about tan graphs, horrible things.
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16 years ago
#33
(Original post by 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?
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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16 years ago
#34
(Original post by 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.
Can you explain it again? I'm getting confused It seems as though you are using a circular argument.
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16 years ago
#35
(Original post by sazzles)
No problem. I agree with you about tan graphs, horrible things.
I never drew graphs for those types of questions. We were taught to draw a quadrant diagram and used that.
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16 years ago
#36
(Original post by sazzles)
Can you explain it again? I'm getting confused It seems as though you are using a circular argument.
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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16 years ago
#37
(Original post by 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.
I'm probably being really stupid here, but with your arguement isn't there the possibility of having a sector with a different angle, that has a larger area than the square. I thought I had phrased the question so it was prove that with a set perimeter, the maximum area of a sector is equal to the area of a square. I still can't see how your logic proves this (try using one syllable words - I'm kind of dumb).
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16 years ago
#38
Ok here's a pretty easy but quite fun problem - find digits a,b,c,d,e such that 4(abcde) = edcba
0
#39
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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16 years ago
#40
those tan graphs came up in the exam 2day and i knew how 2 do them! thanks 4 the help
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