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: $$(\sin {x}\times \sin {x}\times \cos {x})^2+(\cos {x}\times \cos {x}\times \sin {x})^2=(\sin {x}\times \cos {x})^2$$ 
$$\frac{(\sin {x}\times \sin {x}\times \cos {x})^2}{(\sin {x}\times \cos {x})^2}=(\sin{x})^2$$
$$\frac{(\cos {x}\times \cos {x}\times \sin {x})^2}{(\sin {x}\times \cos {x})^2}=(\cos{x})^2$$
 A: 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.
A: Think about similar triangles.  Two similar triangles have exactly the same angles, but the sides are (generally) not the same length.  That fact alone tells you that it is not possible to determine the lengths of the sides of a triangle if all you know is the angles -- you have to also know at least one side length in order to fix the scale.
A: You have:
$$(\text{Hypotenuse,side,side})=(x\sqrt{2},x,x)$$
To see this, use basic trigonometry. 
A: 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. 
