Re: solving for 2 angles

*To*: mathgroup at smc.vnet.net*Subject*: [mg89826] Re: solving for 2 angles*From*: "Dr. Wolfgang Hintze" <weh at snafu.de>*Date*: Sat, 21 Jun 2008 05:33:22 -0400 (EDT)*References*: <g3g00h$kvd$1@smc.vnet.net>

"Steve" <srjm72499 at gmail.com> schrieb im Newsbeitrag news:g3g00h$kvd$1 at smc.vnet.net... >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 > You should write the expressions in Mathematica syntax. Then the equations eq1 = J*Cos[az]*Sin[el] + K*Sin[az]*Sin[el] + L*Cos[el] == 0; eq2 = M*Cos[az]*Sin[el] + N*Sin[az]*Sin[el] + P*Cos[el] == 0; are solved like this sol = Simplify[Solve[eq1 && eq2, {az, el}]]; (dropping here the lengthy expressions for el -> ArcCos[...], az -> ArcCos[...] containing J, K, .., P) The result consists of 16 solutions Length[sol] 16 Regards, Wolfgang