In a $\triangle ABC$, if $a = 13$, $b = 4$ and $\cos C = - \dfrac{5}{13}$, find $R$, $r$, $r_1$, $r_2$ and $r_3$.
Given: $\;$ In $\triangle ABC$, $\;$ $a = 13$, $\;$ $b = 4$, $\;$ $\cos C = - \dfrac{5}{13}$
By cosine rule, $\;$ $\cos C = \dfrac{a^2 + b^2 - c^2}{2 a b}$
$\implies$ $c^2 = a^2 + b^2 - 2 a b \cos C$
i.e. $\;$ $c^2 = \left(13\right)^2 + \left(4\right)^2 - 2 \times 13 \times 4 \times \left(- \dfrac{5}{13}\right)$
i.e. $\;$ $c^2 = 169 + 16 + 40 = 225$ $\implies$ $c = 15$
Also, $\;$ $\sin C = \sqrt{1 - \cos^2 C}$
i.e. $\;$ $\sin C = \sqrt{1 - \left(- \dfrac{5}{13}\right)^2} = \sqrt{\dfrac{144}{169}} = \dfrac{12}{13}$
Now, circumradius $\;$ $R = \dfrac{c}{\sin C} = \dfrac{15}{2 \times \dfrac{12}{13}} = 8.125$
Semi-perimeter of $\triangle ABC$ $\;$ $= s = \dfrac{a + b + c}{2} = \dfrac{13 + 4 + 15}{2} = 16$
Area of $\triangle ABC$ $\;$ $= \Delta = \dfrac{1}{2}a b \sin C = \dfrac{1}{2} \times 13 \times 4 \times \dfrac{12}{13} = 24$
Now, inradius $\;$ $r = \dfrac{\Delta}{s} = \dfrac{24}{16} = 1.5$
Escribed radius $\;$ $r_1 = \dfrac{\Delta}{s - a} = \dfrac{24}{16 - 13} = 8$
Escribed radius $\;$ $r_2 = \dfrac{\Delta}{s - b} = \dfrac{24}{16 - 4} = 2$
Escribed radius $\;$ $r_3 = \dfrac{\Delta}{s - c} = \dfrac{24}{16 - 15} = 24$