Finding the lengths of the sides of a triangle given 3 angles only. If a right $\triangle ABC$ has $\angle A= 90^\circ$, $\angle B=45^\circ$, $\angle C=45^\circ$. Is there a way of finding the lengths of the sides $a$, $b$ & $c$ without knowing any of their lengths ? Normally we use $(\cos{x})^2+(\sin{x})^2=1$ as the hypotenuse.
I have an equation that says: $$(\sin {x}\times \sin {x}\times \cos {x})^2+(\cos {x}\times \cos {x}\times \sin {x})^2=(\sin {x}\times \cos {x})^2$$ 
$$\frac{(\sin {x}\times \sin {x}\times \cos {x})^2}{(\sin {x}\times \cos {x})^2}=(\sin{x})^2$$
$$\frac{(\cos {x}\times \cos {x}\times \sin {x})^2}{(\sin {x}\times \cos {x})^2}=(\cos{x})^2$$

Proof:if x,y and z belong to real numbers and$x<y<z$  then
$(((\frac{\sqrt\frac{y}{z}}{(1-\frac{x}{z})\times\sqrt\frac{x+z}{z-x}})\times\frac{x}{z})+\sqrt\frac{z-y}{z})\times((1-\frac{x}{z})\times\sqrt\frac{(x+z)}{(z-x)})=\sin A  \tag{1}$
$(\frac{\sqrt\frac{y}{z}}{(1-\frac{x}{z})\times\sqrt\frac{x+z}{z-x}})-(((\frac{\sqrt\frac{y}{z}}{(1-\frac{x}{z})\times\sqrt\frac{x+z}{z-x}})\times\frac{x}{z})+\sqrt\frac{z-y}{z})\times(\frac{x}{z})=\cos A \tag{2}$
$\sqrt\frac{(z-y)}{z}=\cos B \tag{3}$
$\sqrt\frac{y}{z}=\sin B \tag{4}$
$\frac{x}{z}=\cos C \tag{5}$
$((1-\frac{x}{z})\times\sqrt\frac{(z+x)}{(z-x)})=\sin C \tag{6}$
$(\sqrt{\frac{y}{z}}\times\frac{x}{z})+\sqrt\frac{z-y}{z}\times((1-\frac{x}{z})\times\sqrt\frac{(z+x)}{(z-x)})=\sin A \tag{7}$
$(-\sqrt{\frac{z-y}{z}})\times\frac{x}{z})+\sqrt{\frac{y}{z}}\times((1-\frac{x}{z})\times\sqrt\frac{(z+x)}{(z-x)})=\cos A \tag{8}$
$((((\frac{\sin B}{\sin C})\cos C)+\cos B)\sin C)=\sin A \tag{9}$
$(\frac{\sin B}{\sin C})-((((\frac{\sin B}{\sin C})\cos C)+\cos B)\cos C)=\cos A \tag{10}$
$((((\frac{\sin A}{\sin C})\cos C)+\cos A)\sin C)=\sin B \tag{11}$
$(\frac{\sin A}{\sin C})-((((\frac{\sin A}{\sin C})\cos C)+\cos A)\cos C)=\cos B \tag{12}$
find $\cos C$ and $\sin C$ in terms of $\sin A, \cos A, \sin B, \cos B$ just like equations $(9)$ to $(12)$?
Notice  that the last four lines, equations $(9)$ to $(12)$, are symmetric with respect to $A,B,C.$ The missing equations are
$ ((((\frac{\sin B}{\sin A})\cos A)+\cos B)\sin A) = \sin C, \tag{13}$
$ (\frac{\sin B}{\sin A})-((((\frac{\sin B}{\sin A})\cos A)+\cos B)\cos A) = \cos C. \tag{14}$ 
The reason is that $\, A + B + C = \pi \,$ because they are angles of a triangle. As just one example, let $\, x=1, \, y=2, \, z=4. \,$ Then the angles are
 $\, A \approx 59.4775^\circ, \, B = 45^\circ, \, C \approx 75.5224^\circ. \,$
The unusual parametrization of the triangle, given $\, x,y,z \,$ and $\, R, \,$ the radius of the circumscribed circle, is 
$$\, b = 2R \sin B.  \sin B=\sqrt{y/z}, \quad c = 2R \sin C. \sin C=\sqrt{1 - x^2/z^2},
 \quad a = 2R \sin A = b \cos C + c \cos B. \,$$
Assuming A,B,C form a triangle.


*

*cos(A+B+C)=-1, cos(A+B)cos(C)-sin(A+B)sin(C)=-1.

*sin(A+B+C)=0, sin(A+B)cos(C)+cos(A+B)sin(C)=0.

*sin(A+B)=sin(A)cos(B)+cos(A)sin(B).

*cos(A+B)=cos(A)cos(B)-sin(A)sin(B).


Knowing cos(A) etc., use 3. and 4. to get cos(A+B) and sin(A+B).
1. and 2. become simultaneous linear equations in cos(C) and sin(C) - readily solved. 
