Question

The sides of a triangle ABC are 6, 7, 8, the smallest angle being C, then the length of internal bisector of angle C is ...

Answer 1

$\frac{14}{5}\sqrt{6}$
If x be the length of internal bisector of $\angle$ C, then it divides the opposite side in the ratio of arms of the angle.
$\therefore \frac{AD}{8}=\frac{BD}{7}=\frac{AD+BD}{8+7}=\frac{6}{15}=\frac{2}{5}$ $\therefore AD=\frac{16}{5}$
Applying cosine formula on $\Delta ADC$,
$A{D}^2=x^2+8^2-2.x.8cos\frac{C}{2}$
${\left(\frac{16}{5}\right)}^2=x^2+64-16x.cos\frac{C}{2}$ ...(1)
where $cos\frac{C}{2}=\sqrt{\frac{s(s-c)}{ab}}=\frac{3}{4}\sqrt{\frac{3}{2}}$
Hence from (1) on putting for $cos\frac{C}{2},$ we get $x=\frac{14}{5}\sqrt{6}$