How do you find all three angles ?Are there any proof of how to claim all three?

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For a triangle of $\angle A B C$ the sides of $a,b,c$ are written in a way of $a=\frac{\sin A}{\sin C}$, $b=\frac{\sin B}{\sin C}$, $c=\frac{\sin C}{\sin C}$ and the heights $h_a,h_b,h_c$ are written in a form $\frac {h_c}{h_a}=$,$\frac{h_c}{h_b}=$,$\frac{h_c}{h_c}$ to give us $a$ and $b$. and the base $c=1$.

If I have a triangle with sides $5,5,4$ and their altitudes are $\sqrt {21},\sqrt {13.44},\sqrt {13.44}$ , why do they simplify to give us a special kind of triangle where $a=1.25,b=1.25,c=1$

$\sqrt\frac {21}{13.44}=1.25$

Angles $3=0.16+0.16+0.68^2+0.84+0.84+(1-0.68^2)$

Laws of Cosine ,when we have all $3$ lengths are:

$a^2=b^2+c^2-2bc\cos(A)$

$b^2=a^2+c^2-2ac\cos(B)$

$c^2=a^2+b^2-2ab\cos(C)$

Here we have sides $1.25,1.25 and 1$,a simplest version of the triangle measuring $5,5,4$.

And for three sides of a triangle $a,b,c$,and $\angle ABC$. The legs of the heights $h_a,h_b,h_c$ are situated on three sides of the triangle. For one side $a$ I have $\frac{b-\cos(A)}{\cos(C)}=a$ and for the second side $b$ I have $\frac{a-\cos(B)}{\cos(C)}=b$ and the third side which is $c$ as the base of the triangle equals to $1$.

$\sqrt{\sin^2(B)+(a-\cos(B))^2}=b$

$\sqrt{\sin^2(A)+(b-\cos(A))^2}=a$

$\sqrt{\sin^2(A)+(\cos(A))^2}=c$

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$\frac{a}{c}=\cos(C)$ $(1-\frac{a}{c})\times\sqrt\frac{(a+c)}{(c-a)}=\sin(C)$

$\sqrt\frac{(c-b)}{c}=\cos(B)$ $\sqrt\frac{b}{c}=\sin(B)$

Find

$\cos(A)$ & $\sin(A)$ which we already know.

Question is: why $\frac{h_c}{h_b}$; $\frac{h_c}{h_a}$;$\frac{h_c}{h_c}$ and. $\frac{\sin A}{\sin C}$; $\frac{\sin B}{\sin C}$ $\frac{\sin C}{\sin C}$equal to the sides $a,b,c$?