When the triangle is constant (meaning that the base equals to one),the sum of the angles in degrees are 180

Can you mix dimensionless function or angles with lenght in triangles?For example two sides are composed of a distance of $0.85+0.4=1.25$ and at the same time $0.4=\cos\theta$and the base is $1$?

For consecutive number or non consecutive numbers $x<y<z$ I have the following example:

$(((\frac{\sqrt\frac{y}{z}}{(1-\frac{x}{z})\times\sqrt\frac{x+z}{z-x}})\times\frac{x}{z})+\sqrt\frac{z-y}{z})\times((1-\frac{x}{z})\times\sqrt\frac{(x+z)}{(z-x)})=\sin A$

$(\frac{\sqrt\frac{y}{z}}{(1-\frac{x}{z})\times\sqrt\frac{x+z}{z-x}})-(((\frac{\sqrt\frac{y}{z}}{(1-\frac{x}{z})\times\sqrt\frac{x+z}{z-x}})\times\frac{x}{z})+\sqrt\frac{z-y}{z})\times(\frac{x}{z})=\cos A$

$\sqrt\frac{(z-y)}{z}=\cos B$

$\sqrt\frac{y}{z}=\sin B$

$\frac{x}{z}=\cos C$

$((1-\frac{x}{z})\times\sqrt\frac{(z+x)}{(z-x)})=\sin C$

$(\sqrt{\frac{y}{z}}\times\frac{x}{z})+\sqrt\frac{ z-y}{z}\times((1-\frac{x}{z})\times\sqrt\frac{(z+x)}{(z-x)})=\sin A$

$(-\sqrt{\frac{z-y}{z}})\times\frac{x}{z}+\sqrt{\frac{y}{z}}\times( (1-\frac{x}{z})\times\sqrt\frac{(z+x)}{(z-x)})=\cos A$

The following variables $a,b,c$ represent the length of the sides of the triangles.The angles have no unit but the lengths do.

$\frac{\sin A}{\sin C}=a$

$\frac{\sin B}{\sin C}=b$

$\frac{\sin C}{\sin C}=c$

h=altitude $\frac{h_c}{h_a}=a$

$\frac{h_c}{h_b}=b$

$\frac{h_c}{h_c}=c$

$((((\frac{\sin B}{\sin C})\times\cos C)+\cos B)\times\sin C)=\sin A$

$(\frac{\sin B}{\sin C})-((((\frac{\sin B}{\sin C})\times\cos C)+\cos B)\times\cos C)=\cos A$

$((((\frac{\sin A}{\sin C})\times\cos C)+\cos A)\times\sin C)=\sin B$

$(\frac{\sin A}{\sin C})-((((\frac{\sin A}{\sin C})\times\cos C)+\cos A)\times\cos C)=\cos B$

And you can obtain all six theta of trigonometric functions with consecutive numbers or non consecutive where $x<y<z$.