Left Hand Wall

$x=-y\ cot\ 29.5^{\circ}+0.09156$

Parameterizing this

$r\mapsto {-y\ cot\ 29.5^{\circ}+0.09156,y,0}$ $t\mapsto {t\ cos\ 29.5^{\circ}+0.09156,-t\ sin\ 29.5^{\circ},0}$

where the negative sign is applied to the sine function by the even odd relationships of cosine and sine, i.e. ( sin(-t)=-sin(t), cos(-t)=cos(t)) and the fact that the y component is in the 4th quadrant.

(x y z

)=(cos 6\[Degree] -sin 6\[Degree] 0 sin 6\[Degree] cos 6\[Degree] 0 0 0 1

) . (x' y' z'

)

(x y z

)=(cos 6\[Degree] -sin 6\[Degree] 0 sin 6\[Degree] cos 6\[Degree] 0 0 0 1

) . (t cos (29.5\[Degree])+0.09156 -t sin (29.5\[Degree]) 0

)

(x y z

)= (0.09156cos 6 \[Degree]+t cos 6 \[Degree]cos (29.5\[Degree])+t sin 6 \[Degree]sin (29.5\[Degree]) -t cos 6 \[Degree]sin (29.5\[Degree])+0.09156 sin 6 \[Degree]+t cos (29.5\[Degree])sin 6 \[Degree] 0

)

(x y z

)= (0.09156cos 6 \[Degree]+t (cos 6 \[Degree]cos(29.5\[Degree])+ sin 6 \[Degree]sin (29.5\[Degree])) 0.09156 sin 6 \[Degree]-t (cos 6 \[Degree]sin (29.5\[Degree])-sin 6 \[Degree] cos (29.5\[Degree])) 0

)

(x y z

)= (0.09156cos 6 \[Degree]+t cos (6\[Degree] -29.5\[Degree]) 0.09156 sin 6 \[Degree]+t sin (6 \[Degree]-29.5\[Degree]) 0

)

(x y z

)= (0.09156cos 6 \[Degree]+t cos (-23.5\[Degree]) 0.09156 sin 6 \[Degree]+t sin (-23.5\[Degree]) 0

)

(x y z

)= (0.09156cos 6 \[Degree]+t cos (23.5\[Degree]) 0.09156 sin 6 \[Degree]-t sin (-23.5\[Degree]) 0

)

Using the equation for y we can solve for t

$y''=0.09156\ sin\ 6^{\circ}-t\ sin\ 23.5^{\circ} \Rightarrow t=\frac{-(y''-0.09156 sin 6^{\circ})}{sin 23.5^{\circ}}$

Substituting this into the expression for x

$x''=0.09156\ cos\ 6^{\circ}+t\ cos\ 23.5^{\circ}$ $x''=0.09156\ cos\ 6 ^{\circ}+\frac{-(y''-0.09156 sin 6 ^{\circ})}{sin 23.5^{\circ}}(cos 23.5^{\circ})$ $x''=0.091058+\frac{y''-.0095706 }{-0.398749} (.917060)$ $x''=0.091058+(y''-.0095706 ) (-2.299843)$ $x''=-2.299843\ y''+.022011+.091058$ $x''=-2.299843\ y''+.113069$