<s>[INST] Can a triangle ABC be made if $\frac{\cos A}{2}=\frac{\cos B}{3}=\frac{\cos C}{7}$ 
I would like to know the simplest approach to find out whether a triangle ABC will be made if $$\frac{\cos A}{2}=\frac{\cos B}{3}=\frac{\cos C}{7}$$

The counterpart questions for sine and tangent can be handled as follows:

*

*If $\dfrac{\sin A}{2}=\dfrac{\sin B}{3}=\dfrac{\sin C}{7}$, we can rule out triangle because by the Sine Rule $a=2k$, $b=3k$, $c=7k \implies a+b <c.$


*If $\dfrac{\tan A}{2}=\dfrac{\tan B}{3}=\dfrac{\tan C}{7}$, we can see that a triangle will be made as $\tan A=2k, \tan B =3k,\tan C=7k$, when inserted in the identity $\tan A+ \tan B+ \tan C= \tan A \tan B \tan C \implies k=\sqrt{2/7}$.
 [/INST] Show for yourself that if $A,B,C$ are the angles of a triangle then
$$
\cos^2A + \cos^2B + \cos^2C = 1-2\cos A \cos B \cos C
$$
This is not very difficult, use the fact that $A+B+C = 180$ along with double angle identities.

Therefore, if each of those ratios equaled $k$ , we get $62k^2 = 1-84k^3$, which can be solved using Cardano's formula (or you can use IVT to just assert the existence of a root) to get you a root like $k \approx 0.117928$. From here, you get that such a triangle in fact exists, and roughly has angles $76.36,69.28$ and $34.36$.
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