Question
In a $$\vartriangle ABC,\cos A = \frac{3}{5}$$ and $$\cos B = \frac{5}{{13}}.$$ The value of $$\cos C$$ can be
A.
$$\frac{7}{{13}}$$
B.
$$\frac{12}{{13}}$$
C.
$$\frac{33}{{65}}$$
D.
None of these
Answer :
$$\frac{33}{{65}}$$
Solution :
$$\tan A = \frac{4}{3}$$ and $$\tan B = \frac{12}{5}.$$ Clearly, $$\tan C$$ should be such that $$\tan A + \tan B + \tan C = \tan A\tan B\tan C$$
$$\eqalign{
& \therefore \,\,\frac{4}{3} + \frac{{12}}{5} + \tan C = \frac{4}{3} \cdot \frac{{12}}{5} \cdot \tan C\,\,\,{\text{or, }}\frac{{56}}{{15}} + \tan C = \frac{{16}}{5}\tan C \cr
& {\text{or, }}\tan C = \frac{{56}}{{33}} \cr
& \therefore \,\,\cos C = \frac{{33}}{{65}}. \cr} $$