Question

In a $△ABC$, if $a=13,b=4,\cos C=-\frac{5}{13}$, then

• A. $r=\frac{3}{2}$
• B. $r_1=8$
• C. $r_2=4$
• D. $r_3=12$

Answer 1

Answer: (A), (B)

$r_1=8$
$r=\frac{3}{2}$

$a=13,b=4,\cos C=-\frac{5}{13}$

By cosine formula, $\cos C=\frac{a^2+b^2-c^2}{2ab}\Rightarrow -\frac{5}{13}=\frac{169+16-c^2}{2ab}$

$\Rightarrow -40=185-c^2\Rightarrow c^2=225\Rightarrow c=15$

Now $\sin C=\sqrt{1-\frac{25}{169}}=\frac{12}{13}$

$\therefore \Delta =\frac{1}{2}ab\sin C=\frac{1}{2}\left(13\right)\left(4\right)\left(\frac{12}{13}\right)=24$

Now $R=\frac{abc}{4\Delta }=\frac{\left(13\right)\left(4\right)\left(15\right)}{4\left(24\right)}=\frac{65}{8}$

$r=\frac{\Delta }{s}=\frac{24\times 2}{a+b+c}=\frac{48}{13+4+15}=\frac{3}{2}$

$r_1=\frac{\Delta }{s-a}=\frac{2\Delta }{(b+c-a)}=\frac{48}{4+15-13}=8$

$r_2=\frac{\Delta }{s-b}=\frac{2\Delta }{a+c-b}=\frac{48}{13+15-4}=2$

$r_3=\frac{\Delta }{s-c}=\frac{2\Delta }{a+b-c}=\frac{48}{13+4-15}=24$