geometry

Using the half-angle formulae:

                                                        \(tan(A)=\frac{2tan(\frac{A}{2})}{1-tan^2(\frac{A}{2})}=\frac{2(\frac{1}{4})}{1-(\frac{1}{4})^2}=\frac{8}{15}\)  .........(1)

                                                        \(tan(B)=\frac{2tan(\frac{B}{2})}{1-tan^2(\frac{B}{2})}=\frac{2(\frac{1}{5})^2}{1-(\frac{1}{5})^2}=\frac{5}{12}\) ..........(2)

                                                                   \(tan(C)=\frac{2tan(\frac{C}{2})}{1-tan^2(\frac{C}{2})}\)                ........(*)

Since the angles in a triangle sum to 180 degrees, then:

                                              \(A+B+C=180 \iff tan(A+B+C)=0\)  (Take tan of both sides).

 

Thus, consider:                        \(tan((A+B)+C)=\frac{tan(A+B)+tan(C)}{1-tan(A+B)tan(C)}=0\)        ........(**)

Now consider \(tan(A+B)\), and using (1) and (2) gives:

                                             \(tan(A+B)=\frac{tan(A)+tan(B}{1-tan(A)tan(B)}=\frac{\frac{8}{15}+\frac{5}{12}}{1-(\frac{8}{15}*\frac{5}{12})}=\frac{171}{140}\)      ......(***)

Thus, substituting (***) in (**) gives:

                                                \(\frac{171}{140}+tan(C)=0 \iff tan(C)=-\frac{171}{140}\)

 

Next, consider (*) and let  \(t=tan(\frac{C}{2})\)

Then, 

                                            \(-\frac{171}{140}=\frac{2t}{1-t^2} \iff 171t^2-171=280t \)

Then,                              \( 171t^2-280t-171=0 \iff(19t+9)(9t-19)=0\)

Thus,                                                       \(t_1=\frac{19}{9},t_2=-\frac{9}{19}\)

But, C is a positive angle (Angle in a triangle), hence \(t=tan(\frac{C}{2})>0\)

Therefore, only \(t_1=\frac{19}{9}\) is the desired solution.

Thus,   \(tan(\frac{C}{2})=\frac{19}{9}\)

 

Hope this helps, and I think there is a simpler solution than this!