l := 89: h := 49: d:= 55: beta1 := 11.5*Pi/180:

A := l*sin(beta1): B := l*cos(beta1): C := (h + 0.5*d)*sin(beta1) - 0.5*d*tan(beta1):

E := (h + 0.5*d)*cos(beta1) - 0.5*d:

RootFinding[Analytic](A*sin(alpha)*cos(alpha) + B*sin(2*alpha) - C*cos(alpha) - E*sin(alpha) = 0, alpha, re=0..2*Pi, im=-1..1);

evalf(map(convert,[%], degrees));

There is no  33 degrees or  22 degrees  in this list .

Your formula for the width of the wall of the players can be simplified since

sqrt(6400*x^4+12800*x^2*y^2+6400*y^4) = 80*(x^2+y^2)

and needs some refinement. This formula is true if it is assumed that the width of a football goal is 8 meters, and the distance from the point  (x, y)  to the wall is 10 meters. In fact, the gate width should be equal to 7.32 meters, and the distance to the wall  9 meters. 

With these adjustments more accurate formula:

z = (65.88*(x^2+y^2))*y/abs((x^2+y^2-3.66*x)*(3.66*x+x^2+y^2))

We construct a plot of this function, taking into account that the domain is a football field minus the penalty area:

restart;

z := unapply(piecewise(abs(x)<20.15 and y>0 and y<16.5, undefined, (65.88*(x^2+y^2)*y)/abs((x^2+y^2-3.66*x)*(3.66*x+x^2+y^2))), x, y):

plot3d(z, -34..34, 0..40, view=[-34..34, 0..40, 0..4.5], style=surface, axes=normal, numpoints=10000, orientation=[-40, 75], lightmodel=light4, labels=[x, y, z]);

plot3d([27*(t-sin(t)), sqrt((27*(1-cos(t)))^2)*cos(s), sqrt((27*(1-cos(t)))^2)*sin(s)], t=0..2*Pi, s=0..2*Pi);

ListTools[Reverse]([[1,4],[2,3],[3,2],[4,1],[6,5],[6,1]]);

                         [[6, 1], [6, 5], [4, 1], [3, 2], [2, 3], [1, 4]]

a[1], a[2], a[3]:=3, -1, 2:

for n from 3 to 89 do

a[n+1]:=a[n]-3*a[n-1]+a[n-2]:

od:

a[90];

                   58692136618345153331

A:=-9-60*s1/(60*s1-1)+500*s1^2/(10*s1-1)/(60*s1^2/(10*s1-1)-1):

solve({A>=2.5, A<5});

            {0.01531863559 <= s1, s1 < 0.01554503851}, {0.1400885995 <= s1, s1 <= 0.3572594316}

Procedure  Rename  solves your problem. If solutions of the system depend on a single parameter, it is denoted by  t , and if solutions depend on  r  parameters, they are denoted by  t[1], t[2], ... , t[r] .

Rename:=proc(sys)

local Ind, S, L, n, r;

Ind:=indets(sys);

n:=nops(Ind);

solve(sys);

L:=[seq(rhs(%[i]), i=1..n)];

S:=indets({op(L)});

r:=nops(S);

if r=1 then {seq(Ind[i]=eval(L[i],{op(S)=t}), i=1..n)} else

{seq(Ind[i]=eval(L[i],{seq(S[j]=t[j], j=1..r)}), i=1..n)} fi;

end proc;

Example:

sys:={x[1]-x[2]=1, x[1]+x[2]-x[3]-x[4]=5};

solve(sys);

Rename(sys);

The problem will disappear if you write:

f := x->piecewise(x <= 1, a*x^2+b, 1 <= x, 1/x):

limit(diff(f(x), x), x = 1, right);

limit(diff(f(x), x), x = 1, left);

                                 -1

                                 2 a

Cause of error - the unknown value of the function  f(1)  to calculate the right derivative  at  x=1 .

I do not understand the meaning of the problem. A curve is a set of points, depending on one parameter. In your expressions, there are several parameters. 

with(LinearAlgebra):

add(add(A[i,j]*B[i,j], i=1..RowDimension(A)), j=1..ColumnDimension(A));

S1 := seq(plot3d(0, x = -11 .. h, y = 0 .. sqrt(121-x^2), style = surface), h = -11 .. 11, .2):

S2 := seq(plot3d([x, y, y*sqrt(3)], x = -11 .. h, y = 0 .. (1/2)*sqrt(121-x^2), style = surface), h = -11 .. 11, .2):

S3 := seq(plot3d([x, y, (sqrt(121-x^2)-y)*sqrt(3)], x = -11 .. h, y = (1/2)*sqrt(121-x^2) .. sqrt(121-x^2), style = surface), h = -11 .. 11, .2):

S4 := seq(plottools[polygon]([[x, 0, 0], [x, (1/2)*sqrt(121-x^2), (1/2)*sqrt(121-x^2)*sqrt(3)], [x, sqrt(121-x^2), 0]], color = pink), x = -11 .. 11, .2): C := plots[spacecurve]([11*cos(t), 11*sin(t), 0], t = 0 .. Pi, color = black):

S := seq(plots[display](C, S1[k], S2[k], S3[k], S4[k]), k = 1 .. 111):

plots[display](S, insequence = true, scaling = constrained, axes = normal, lightmodel = light4, view = [-12.5 .. 12.5, -2.5 .. 12.5, -1.5 .. 11.5], orientation = [-12, 67]);

restart;

L:=[]: Set:={$0..9}:

for H in {1,2,3} do

for U in Set minus {H} do

for N in Set minus {H,U} do

for S in Set minus {0,H,U,N} do

for E in Set minus {H,U,N,S} do

for T in Set minus {H,U,N,S,E} do

for R in Set minus {H,U,N,S,E,T} do

for o in Set minus {H,U,N,S,E,T,R} do

for G in Set minus {H,U,N,S,E,T,R,o} do

a:=H*10^5+U*10^4+N*10^3+S*10^2+E*10+N: b:=S*10^5+T*10^4+R*10^3+o*10^2+N*10+G:

if a*3=b then L:=[op(L), [HUNSEN=a, STRONG=b]] fi:

od: od: od: od: od: od: od: od: od:

            [[HUNSEN = 126356, STRONG = 379068], [HUNSEN = 318928, STRONG = 956784]]

The method of solution is similar to one in  http://www.mapleprimes.com/questions/151971-Number-Theory

Obviously, the vertices of the alternating triangle are

A := [x, 0, 0]: B := [x, sqrt(121-x^2), 0]: C := [x, (1/2)*sqrt(121-x^2), (1/2)*sqrt(121-x^2)*sqrt(3)]: 

The code for plotting:

S1 := plot3d(0, x = -11 .. 11, y = 0 .. sqrt(121-x^2), style = surface):  # Lower face of the solid 

S2 := plot3d([x, y, y*sqrt(3)], x = -11 .. 11, y = 0 .. (1/2)*sqrt(121-x^2), style = surface):  # Left face of the solid

S3 := plot3d([x, y, (sqrt(121-x^2)-y)*sqrt(3)], x = -11 .. 11, y = (1/2)*sqrt(121-x^2) .. sqrt(121-x^2), style = surface):  #Right face of the solid

plots[display](S1, S2, S3, scaling = constrained, axes = normal, lightmodel = light4, view = [-12.5 .. 12.5, -2.5 .. 12.5, -1.5 .. 11.5], orientation = [-12, 67]);

Now your body is closed from all sides:

A := plot3d([2*t^3-t^4, (t+2)*cos(theta)-2, (t+2)*sin(theta)], t = 0 .. 2, theta = 0 .. 2*Pi):

B := plots:-spacecurve([t, -2, 0], t = -1 .. 3, color = red, thickness = 2):

C := plot3d([0, -2+r*cos(phi), r*sin(phi)], r = 2 .. 4, phi = 0 .. 2*Pi):

plots:-display(A, B, C, axes = normal, scaling=constrained, view = [-1.4 .. 5.4, -7.4 .. 3.4, -4.4 .. 4.4]);

Code of  Preben Alsholm  was used.

Without plots and plottools:

piecewise(seq(op([x>-7+2*n and x<-7+2*(n+1), exp(-x-6+2*n)]), n=0..7)):

plot(%, x=-7..9, -1..3.5, thickness=2, scaling=constrained, discont=true);