For a triangle of $ \angle A B C$ the sides of triangles $ 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}=$ 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$ , they simplify and 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)$