I'm sure if you go through it step-by-step, or w/o assigning numbers until the end, you'll see that value of b1 is magical and all of the terms in r1v go away
l1=3;
h1=1.45;
syms l1 h1 real
b1=sqrt(l1^2+h1^2);
z=atand(h1/l1);
q1 = 0;
syms r1v x
R2v(r1v) =r1v/3 - 209/1395;
Mz1v(r1v,x) = -r1v + (R2v(r1v) * cosd(z)) * x + (q1 * sind(z)) * x
Mz1v = simplify(Mz1v)
Mz1v(r1v,b1)
If l1 = 3, as in this case, then the r1v is annihilated, and we're left with
subs(ans,l1,3)



