Is $\cos^2A+\cos^2B+\cos^2C+2 \cos A \cos B \cos C=1$ both Necessary and Sufficient for a triangle

Question

When $P^2+Q^2+R^2+2PQR=1$, one can show that all of $|P|,|Q|,|R|$ are either $\ge 1$ or $\le 1$. One of the consequences of this is the trigonometric identity in a triangle: $$\cos^2 A+ \cos^2 B+ \cos^2 C+2 \cos A \cos B \cos C=1.$$ We want to point out that it is only necessary but not sufficient for forming a triangle. Given $\cos A, \cos B$, we can set up the quadratic $$\cos^2 C + (2\cos A \cos B) \cos C+ \cos^2 A+ \cos^2 B-1=0,\quad (*)$$ We get two roots for $C: C_1, C_2$, then either ABC$_1$ or ABC$_2$, will be a triangle such that $A+B+C=\pi$. So if three numbers $\cos A, \cos B. \cos C$ satisfy (*) a triangle may or may not be formed.

On the other hand the identity $\tan A+ \tan B + \tan C=\tan A \tan B \tan C$ is both necessary and sufficient for the formation of a triangle.

Question no. 1: Check if $\cos A=1/2, \cos B=3/5, \cos C=3/7$ would form a triangle?

Question no. 2: If $\cos A=\frac45, \cos B=\frac37$ and the triangle $ABC$ is possible, find the value of $\cos C$.

Answer 1

Question 1: $\arccos(\frac12)+\arccos(\frac35)+\arccos(\frac37)\approx3.10237805192\ne\pi$. Consequently the answer is NO.

Question 2: $\arccos(\frac45)+\arccos(\frac37)\approx1.77138636151$ so we must have $\arccos(C)\approx\pi-1.77138636151\approx1.37020626208$. From this $$\cos(1.37020626208)\approx0.199247598881 \text{ radians }\approx11.41604649^{\circ}$$ NOTE.- Obviously the two questions can also be answered using the above identity.

Answer 2

The fun is to do these questions by hand without any electronic aid to calculate Arccos(.)

Let $f=\cos^2 A + \cos^2 B+\cos^2 C+2 \cos A \cos B \cos C-1$ by putting the given cos values: 1/2. 3/5.3/7, we get $$f=\frac{1}{4}+\frac{9}{25}+\frac{9}{49}+\frac{9}{35}-1= \approx 0.25+ 0.36+ 0.183+ 0.257-1 \approx 0.0508 \ne 0.$$ So a triangle is not possible.

Next let us put $cos A =4/5. \cos B=3/7$ and $\cos C=c$ in $f=0$, we get a quadratic in $c$. We get $$c^2+\frac{24}{35}c-\frac{216}{1225}=0 \implies c=\cos C=\frac{6}{35}[-2\pm \sqrt{10}].$$ The positive sign goves the correct value.