Is the question: find the value of $ cosA $ with parameter? Or Is the question: find the value of $ cosA $ exact? Examples. If $c=10, a=7.9, b=2.400909$ then $A = 25.46^{o}$ If $c=10, a=8.0, b=2.4060606$ then $A = 29.72^{o}$ If $c=10, a=8.1, b=2.411688$ then $A = 33.53^{o}$ If $c=10, a=8.2, b=2.417779$ then $A = 37.03^{o}$

If B = 90 degrees, cosA = [11 +/- sqrt(41)]/18.

Probably a copy-catter of the problem 2 of Vietnam Mathematical Olympiad 1976: Find all triangles ABC such that $ \frac{acosA+bcosB+ccosC}{asinA+bsinB+csinC}=\frac{a + b + c}{9R} $ Hey, I am good at catching the copy-catters... LOL.

a=2RsinA,b=2RsinB c=2RsinC, R(sin2A+sin2B+sin2C)\2R(sin^2 A+sin^2 B+sin^2 C)=2R(sinA+sinB+sinC)\9R sin2A+sin2B+sin2C=4sinAsinBsinC 9sinAsinBsinC=(sin^2 A+sin^2 B+sin^2 C)(sinA+sinB+sinC)>=9sinAsinBsinC -> a=b=c

PP. Prove that in any triangle $ABC$ there is the implication $\frac {\sum a\cos A}{\sum a\sin A}=\frac {2s}{9R}\implies a=b=c$ . Proof. Prove easily that $\left\{\begin{array}{c} \sum a\cos A=R\sum \sin 2A=4R\prod\sin A=\frac {2S}{R}=\frac {2sr}{R}\\\\ \sum a\sin A=\sum\frac {a^2}{2R}=\frac 1{2R}\cdot \sum a^2=\frac {s^2-r^2-4Rr}{R}\end{array}\right|$ . Therefore, $\frac {\sum a\cos A}{\sum a\sin A}=\frac {2s}{9R}\iff$ $\frac {\frac {2sr}{R}}{\frac {s^2-r^2-4Rr}{R}}=\frac {2s}{9R}\iff$ $s^2-r^2-4Rr=9Rr\iff$ $s^2=r^2+13Rr$ . Since $s^2\ge 16Rr-5r^2$ obtain that $r^2+13Rr\ge 16Rr-5r^2\implies$ $3Rr\le 6r\implies$ $R\le 2r\le R\implies R=2r\implies a=b=c$ .