For a triangle of $\angle A B C$ the sides of $ a,b,c$ are presented as $a=\frac{\sin A}{\sin C}$, $b=\frac{\sin B}{\sin C}$, $c=\frac{\sin C}{\sin C}$ 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 $c=1$.


example:


Assuming I have a triangle with sides $5,5,4 $ and their altitudes $\sqrt 21,\sqrt 13.44,\sqrt 13.44$ , why do they simplify to give us a special kind of triangles?


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


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:


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


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


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


Per example here we have sides 1.25,1.25 and 1,a simplest version of the triangle measuring 5,5,4.


I have 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 equal to 1.




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




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


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




by using consecutive or non consecutive numbers:
$a<b<c $ we're able to define $\theta$ without using $\pi$ radian for Cosine or Sine.


example of consecutive numbers:


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


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


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