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**GurraTedden****Member**- Registered: 2005-07-20
- Posts: 19

This suddenly felt impossible:

How do I transform these three equations...

(1): xz=bc

(2): cos(y) + xsin(y)=0

(3): zsin(y) + xzcos(y) - c(a+b)=0

(three unknown x,y,z with three equations, always solvable right!? But i only want to solve this for x!!!)

(I've looked it through carefully, so it shouldn't be any errors in there)

...to either of these too:

answer, version1: x=1/√((a/b+1)²-1)

, version2: x=tan(arcsin(b/(a+b))

This is how far I've got:

(1) can be rewritten to: x=bc/z , new(4)

(2) -"- : x=cos(y)/sin(y) , new(5)

(3) -"- : z=c(a+b)/(sin(y)+xcos(y)) ,new(6)

(6) in (4) --> x=b(sin(y)+xcos(y))/(a+b) , new(7)

(5) in (7) --> cos(y)/sin(y)=b(sin(y)+cos²(y)/sin(y))/(a+b) --> [multiplying both sides with sin(y)] cos(y)=b/(b+c)

So, finally: I'm getting y=arccos(b/(a+b)), BUT HOW DO I PROCEED... It's a dead end. I don't know what more to do...

Maybe you do?

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**GurraTedden****Member**- Registered: 2005-07-20
- Posts: 19

CORRECTION: eq(2) should be: cos(y) - xsin(y)=0!

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 13,895

GurraTedden wrote:

(three unknown x,y,z with three equations, always solvable right!? But i only want to solve this for x!!!)

There are actually six variables...x,y,z,a,b,c

Character is who you are when no one is looking.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Yes, but he can measure a, b and c, so he wants to make x the subject of an equation involving just those.

Why did the vector cross the road?

It wanted to be normal.

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