Re: solving for 2 angles

• To: mathgroup at smc.vnet.net
• Subject: [mg89821] Re: solving for 2 angles
• From: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
• Date: Sat, 21 Jun 2008 05:32:25 -0400 (EDT)
• Organization: The Open University, Milton Keynes, UK
• References: <g3g00h\$kvd\$1@smc.vnet.net>

```Steve 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)
>
>
> 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.

Steve,

This newsgroup is dedicated to Mathematica, software made and published
by Wolfram Research, Inc. Thus, I assume that you are using it and I
shall show you the syntax and command to use to get the complete set of
solutions.

eqns =
{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};
Reduce[eqns, {az, el}]

((C[1] | C[2]) \[Element] Integers && ((L == 0 && P == 0 &&

az == Pi + 2 Pi C[1] && el == Pi + 2 Pi C[2]) ||

2    2           2    2                L M
(J != 0 && ((L (J  + L  - J Sqrt[J  + L ]) != 0 && P == --- &&
J

az == Pi + 2 Pi C[1] &&

2    2
-J + Sqrt[J  + L ]
el == 2 (ArcTan[------------------] + Pi C[2])) ||
L

2    2           2    2                L M
(L (J  + L  + J Sqrt[J  + L ]) != 0 && P == --- &&
J
<...>

(* the rest of this *very* long output has been deleted for the reader's
sake. Just evaluate the original expressions within Mathematica on your
system to get it again. *)

Regards,
-- Jean-Marc

```

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