# How can i find the solution of the equation when the R.H.S also contains the unknown??

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Naveen Ramesh on 23 Apr 2013
I have this eqaution that need to be plotted between s_y and beta. But the problem is that, the R.H.S also contains the quantity beta but under a different math term. This is the eqation:
beta=acosd((((R^2-((R*u*sin(beta))/(R-d))^2)^0.5*(R-d))/(R*u))-(s_y/u));
Except for Beta, all the values are known and s_y has been defined as
s_y=0:0.1:35;
How can i do this?? meaning, how can i solve for beta? am new to matlab and just learning the basics. So if anybody is able to help me with this,I ll be grateful.....
Naveen Ramesh on 24 Apr 2013
R,d and u are constants and Walter you are right.. !!it was a typo error... It is sind(beta)

Walter Roberson on 23 Apr 2013
atan2(sqrt(-(-d+R-s_y+u) .* (-d+R+s_y-u) .* (-d+R+s_y+u) .* (-d+R-s_y-u)) ./ (u.*s_y), (R.^2-2.*R.*d-u.^2+d.^2-s_y.^2) ./ (u.*s_y))
and also the negative of that.
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Walter Roberson on 24 Apr 2013
There are two solutions. atan2() is the arctan function that requires two arguments in order to properly figure out which quadrant to use, so atan2() is going to return one value. The negative of that value is also a solution for a total of two solutions.