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Re: solving for 2 angles

  • To: mathgroup at smc.vnet.net
  • Subject: [mg89820] Re: [mg89788] solving for 2 angles
  • From: "W_Craig Carter" <ccarter at mit.edu>
  • Date: Sat, 21 Jun 2008 05:32:14 -0400 (EDT)
  • References: <200806201015.GAA21484@smc.vnet.net>

Hello Steve,
Are you asking how to do this in Mathematica?

If so, this gives you *some* solutions, but perhaps not all.

eq1 = jj*Cos[az]*Sin[el] + kk*Sin[az]*Sin[el] + ll*Cos[el]
eq2 = mm*Cos[az]*Sin[el] + nn*Sin[az]*Sin[el] + pp*Cos[el]

Solve[{eq1 == 0, eq2 == 0}, {az, el}]
(*produces a long list of solution-replacement rules, and a warning
about inverse functions, and advice about using Reduce*)

Craig

On Fri, Jun 20, 2008 at 6:15 AM, Steve <srjm72499 at gmail.com> wrote:
> I have two equations with unknown angles az and el:
>
> J*cos(az)*sin(el)+K*sin(az)*sin(el)+L*cos(el)=0;
> M*cos(az)*sin(el)+N*sin(az)*sin(el)+P*cos(el)=0;
>
> Is it a simple matter to solve for the angles?
>
> I tried taking the first equation, and re-writing as:
>
> sin(el)*[J*cos(az)+K*sin(az) ] = -L*cos(el)
>
> which leads to
>
> el= atan(-L / [J*cos(az)+K*sin(az)])
>
> I then subsituted this into the second equation, but was not able to
> isolate the az variable.
>
> I would appreciate any help...
>
> Thanks.
>
> -Steve
>
>



-- 
W. Craig Carter


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