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 $a$, $b$ &amp; $c$ 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: $$(\sin {x}\times \sin {x}\times \cos {x})^2+(\cos {x}\times \cos {x}\times \sin {x})^2=(\sin {x}\times \cos {x})^2$$ 

$$\frac{(\sin {x}\times \sin {x}\times \cos {x})^2}{(\sin {x}\times \cos {x})^2}=(\sin{x})^2$$
$$\frac{(\cos {x}\times \cos {x}\times \sin {x})^2}{(\sin {x}\times \cos {x})^2}=(\cos{x})^2$$
Here's an interesting fact:


  If $A$, $B$, $C$ are angles with $A+B+C=180^\circ$, then $\sin A$, $\sin B$, $\sin C$ are the lengths of the sides of a (not the!) triangle with angles $A$, $B$, $C$.


Specifically, the "$\sin A$-$\sin B$-$\sin C$" triangle is the one inscribed in a circle of diameter $1$; you might consider it a fundamental representative of the family of triangles with angles $A$, $B$, $C$, but it is not the only triangle with those angles. You can get from the representative to any other member of the family by magnifying the side-lengths (and the circumdiameter) by some factor, $m$; conversely, any member of the family has side-lengths that are multiples of the side-lengths of the representative (since the family members are all similar).


  A triangle has angles $A$, $B$, $C$ if and only if its side-lengths are $m\sin A$, $m\sin B$, $m \sin C$ for some (positive) $m$.


This is, in fact, exactly what the Law of Sines tells you:
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = m$$
where $m$ is that scale factor ... and also the diameter of the triangle's circumcircle.



In the case of a right triangle with hypotenuse $1$, recall Thales' Theorem:


  A right angle is inscribed in a semi-circle.


Thus, an hypotenuse-$1$ right triangle is inscribed in a circle whose diameter matches that hypotenuse: in other words,


  A unit-hypotenuse right triangle is the fundamental representative of the "$A$-$B$-$90^\circ$" family.


Such a triangle's sides are indeed $\sin A$, $\sin B$ (which we also call "$\cos A$", since $A$ and $B$ are complementary), and $\sin 90^\circ$ (which, of course, is $1$). Perhaps this is what you were getting-at in your comment to your question.
Proof:if x,y and z belong to real numbers and$x&lt;y&lt;z$  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.