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 $\tan A+ \tan B+ \tan C= \tan A \tan B \tan C \implies k=\sqrt{2/7}$.

Strating from @Teresa Lisbon's answer, the exact results are
$$k=\frac{31}{126} \left(2 \cos \left(\frac{1}{3} \left(2 \pi  n-\cos
   ^{-1}\left(-\frac{17884}{29791}\right)\right)\right)-1\right)\qquad (n=0,1,2)$$ and this gives angles (in degrees) $a=76.358$, $b=69.281$, $c=34.361$.
Using algebra, the problem is very simple since it reduces to the equation
$$a+\cos ^{-1}\left(\frac{3}{2} \cos (a)\right)+\cos ^{-1}\left(\frac{7 }{2} \cos (a)\right)=\pi$$ which has only one solution.
Using series expansion around $a=\frac \pi 2$ gives
$$0=\frac{\pi }{2}+6 \left(a-\frac{\pi }{2}\right)+\frac{55}{8} \left(a-\frac{\pi
   }{2}\right)^3+\frac{4627}{128} \left(a-\frac{\pi
   }{2}\right)^5+O\left(\left(a-\frac{\pi }{2}\right)^7\right)$$ and series reversion leads to
$$a \sim\frac{5 \pi }{12}+\frac{55 \pi ^3}{82944}+\frac{89 \pi ^5}{10616832}\approx 1.33212$$ while the "exact" solution is $a=1.33270$.
