How to find the measure of the sides of all kinds of triangles. Can it be proven?
Three angles sums equal to $180$. Lets keep per example $90$ degrees as the vertex and $60$ and $30$ as angles.
The question is find the lengths of the three sides? And do these for all triangles,whether it's scalene, isosceles or equilateral. examples of lengths of the sides of triangles by using only three angles.
$\frac{\sin(A)}{\sin(C)}=a$
$\frac{\sin(B)}{\sin(C)}=b$
$\frac{\sin(C)}{\sin(C)}=c$
If I have a right triangle where the vertex equal to $90$ and two angles $60$ & $30$ degrees I use the following answer:
$\frac{\sin(60)} {\sin90}=a$
$\frac{\sin 30}{\sin 90}=b $
$\frac{\sin 90}{\sin90}=c$
For consecutive numbers $a<b<c$
The lengths of the sides $a,b,c$ in terms of $x,y,z$
$x=\frac{\sin(180-\arccos\sqrt\frac{(c-b)}{c}-\arccos\frac{a}{c})}{(1-\frac{a}{c})\times\sqrt\frac{(a+c)}{(c-a)}}$
For Lengths of $b=y$
$ y=\frac{\sqrt\frac{b}{c}}{(1-\frac{a}{c})\times\sqrt\frac{a+c}{c-a}}$
For Lengths of $c=z$
$ z=\frac{(1-\frac{a}{c})\times\sqrt\frac{a+c}{c-a}}{(1-\frac{a}{c})\times\sqrt\frac{a+c}{c-a}}$
$\sin((180-\arcsin\sqrt\frac{b}{c}-\arcsin((1-\frac{a}{c})\times\sqrt\frac{a+c}{c-a})))$
$a<b<c$