Question

In a $\Delta ABC,a=5,b=4$ and $\tan \frac{C}{2}=\sqrt{\frac{7}{9}},$ then measure side ${}^{\prime }c{}^{\prime }$

Answer 1

$\cos C=\frac{1-{\tan }^2\frac{C}{2}}{1-{\tan }^2\frac{C}{2}}=\frac{2}{16}=\frac{1}{8}$

Now using cosine rule,

$\cos C=\frac{a^2+b^2-c^2}{2ab}=\frac{41-c^2}{40}=\frac{1}{8}$

$\Rightarrow c^2=41-5=36\therefore c=6$