The set of equations; as far as I can see, cannot be solved explicitly do to the trig. functions.
I rearranged your equations so you only need to solve eq5 and eq6 for y and z.

I randomly made up values for  t, T, a,c,e,f just to show the use of fsolve. 

Hope this helps you.

Edit I re-attached the worksheet because I had a mistake in it.

Download 2024-05-13_A_An_approach.mw

f:=x-> (1-k*x)/(1+x^2)
f := proc (x) options operator, arrow; (1-k*x)/(1+x^2) end proc

slpf:=diff(f(x),x)
                           k      2 (-k x + 1) x
                slp := - ------ - --------------
                          2                 2   
                         x  + 1     / 2    \    
                                    \x  + 1/    

m:=eval(slpf,x=3);# slope of tangent line at x=3

                              2      3 
                         m := -- k - --
                              25     50

f(3)
                            3      1 
                          - -- k + --
                            10     10

line:=unapply(m*x+c,x)
 line := proc (x) options operator, arrow; ((2/25)*k-3/50)*x+c 

    end proc




line(3)=f(3);# x=3 is tangent point
                  6      9          3      1 
                  -- k - -- + c = - -- k + --
                  25     50         10     10

c:=solve(line(3)=f(3),c)
                               27     7 
                        c := - -- k + --
                               50     25

line(x)
                   /2      3 \     27     7 
                   |-- k - --| x - -- k + --
                   \25     50/     50     25

Check under Tools Options Display Typesetting level to see if it is set to extended.  

@dharr  I appear to have the setup working in principal. I created seperate worksheets. One for the Main Package and one for the Sub Package. Both attached.  I have hilighted the differences in both in red. Then I copied both help files across to the correct location. They now list correctly in Help. 

New_RT_Edit_Dadabase_2024.mw

New_UHG_Edit_Dadabase_2024.mw

Interesting problem.  I found this method in the RealDomain help page.

restart;

n:=3;m:=2;
eqx:=x^(n/m)=a;
use RealDomain in (maple_sol:=solve(eqx,[x]))[] end use;  #also tried solve()
F:=map(X->eval(eqx,X),maple_sol);
map(X->evalb(X),F);
                             n := 3

                             m := 2

                               (3/2)    
                       eqx := x      = a

                          [     (2/3)]
                          [x = a     ]

                          F := [a = a]

                             [true]

This is rather similar to the 1st question I asked on MaplePrimes. I just happened to look at it today

Dual sum loop with a condition - MaplePrimes  

Download A_2024-03-15_Sum_with_Condition.mw

This is a way using seq

M := Matrix(4, 4, [[m[1, 1], m[1, 2], m[1, 3], m[1, 4]], 
                   [m[2, 1], m[2, 2], m[2, 3], m[2, 4]],
                   [m[3, 1], m[3, 2], m[3, 3], m[3, 4]],
                   [m[4, 1], m[4, 2], m[4, 3], m[4, 4]]]);

L:=[seq(M[1 .. 4, i], i = 1 .. 4)];

whattype(L[3]);

I am sure there are more direct ways.1st, I changed you numbers just to see if this works more generally. Works with you numbers too

a:=evalf(1119.8*21/101)
                        a := 232.8297030

b:=trunc(a)
                            b := 232

c:=evalf(a-b)
                         c := 0.8297030

d:=evalf(c,2)
                           d := 0.83

b+d
                             232.83

with your numbers
a:=evalf(3*21/100,3);
                           a := 0.630

b:=trunc(a)
                             b := 0

c:=evalf(a-b)
                           c := 0.630

d:=evalf(c,2)
                           d := 0.63

b+d
                              0.63

Your code has so many errors. Are you typing it in a text editor? Hard to see how you could enter that much code in maple and not notice or find errors along the way.

This runs but I cannot verify you logic.

I used the VS code editor to find the errors. and maplemint That might help you.

Cong:=proc(n)
 local  a,b,An,Bn,Cn,Dn:
if n mod 2 = 1  then  
  An:=0:   
  Bn:=0:    
  for a from (round(-sqrt(n/(2)))) to round(sqrt(n/(2)) ) do           
     for b  from (round(-sqrt(n)) )to round(sqrt(n) )do                
          if is( (sqrt(n-2*a^(2)-b^(2)) )/(32),integer  ) then 
               An:=An+1 ;               
             elif is((sqrt(n-2*a^(2)-b^(2)) )/(8) ,integer )  then                   
               Bn:=Bn+1 ;
          end if:          
     end do: 
  end do:
 end if; 
if 2*An=Bn  then
    print(True); 
  else 
   print(False);
end if; 
if n mod 2 = 0 then
    Cn:=0: 
    Dn:=0:
end if;          
for a  from (round(-sqrt(n/(8)))) to round(sqrt(n/(8)) ) do          
    for b  from (round(-sqrt(n/(2)))) to round(sqrt(n/(2))) do                
        if is((sqrt(n/(2)-4*a^(2)-b^(2)) )/(32), integer   ) then                  
              Cn:=Cn+1 ;               
          elif is((sqrt(n/(2)-4*a^(2)-b^(2)) )/(8) ,integer) then
            Dn:=Dn+1 ;
        end if:
    end do: 
end do:  
if 2*Cn=Dn then
    print("True");  
  else  
    print("False");
end if;  
end proc;

maplemint(Cong);
Cong(108);

If I am intrepetering things correctly.  Your

diff((z - zA)*(z - zB)*(z - zC), z)*I);   # I removed semicolin inside brackets

gives a 2nd order polynomial (quadratic)   .say it is Q(Z)    Why have you multiplied the whole diff eqn. by I 

solve(Q(z) , z ,explicit)   #gives two roots which are supposed to be the focii of the ellipse.

Look on Marden's theorem - Wikipedia

This is a really useful procedure I found here on Primes  years ago. Called factrix. though in this case it pulls out (-1/2) instead of (1/2)
Edit:- the link to the reference to factrix  How do I extract common factors in a matrix? - MaplePrimes

[moderator: Copyrighted code removed. Source is factrix by Robert Israel, part of his Maple Advisor Database.]

You have to expand the equation. 

if  expand((x+1)^2=x^2+2*x+1) then print(1) else print(0) fi

or

if  expand((x+1)^2)=x^2+2*x+1 then print(1) else print(0) fi

I parameterised the points on the ellipse. I made them projective points [x,y,1] in the form of vectors. Reason the equation of the  line through two projective points is the Cross Product of the two points as another vector . P1 X P2  = L12 Now the intersect point of two lines as vectors is L1 X L2= Pintersect.

A line in vector form <a , b, c>   represents a*x + b*y + c=0

The use of factrix by @Robert B. Isreal  reduces the projective vectors, as in projective geometry vectors can be scaled without changing the answers.

Edit:- Correction.  Replece this into the document.  The intersection points were scaled incorrectly.

The 3 collinear points

for i to 3 do
    pPint || i := convert(Pint || i[1 .. 2]/Pint || i[3], list);
    eval(pPint || i, pramsplot);
    evalf(%);
end do;

Download 2023-12-23_A_Pascal_Conic_parameterised.mw

You basically have a 5th order polynomail in h. In general there is no explicit solution for a 5th order or higher polynomial.

I asked the same question a couple on months age

@C_R answered it as follows

inserts a new execution group before the cursor.

For pasted text you could try F3 to split the text at the postition of the cursor.

from the menu should also work.