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

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.....

[Image Analyst]

And what's R, d, and u?

[Walter Roberson]

Are you sure? You are using acosd() implying that beta is in degrees, but you use sin(beta) implying that beta is in radians.

[Naveen Ramesh]

R,d and u are constants and Walter you are right.. !!it was a typo error... It is sind(beta)

Answer 1

[Walter Roberson]

Assuming radians,

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.

[Naveen Ramesh]

yu mean there are 4 solutions ??? coz there is a comma in the answer that yu have given and also yu mentioned that the negative of this also forms the answer... ??

[Walter Roberson]

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.