Question

If in $\Delta ABC,r=2,r_1=4,s=12anda<b<c$, then match the following.
The codes for the lists have choices $\left(A\right),\left(B\right),\left(C\right)$ and $\left(D\right)$ out of which only ONE is correct.
Codes :
P Q R S
(A)$4231$
(B)$2143$
(C)$3412$
(D)$2134$

Answer 1

We have;

In $\Delta ABC$;

$r=2=\frac{\Delta }{s}⟹\Delta =2\times 12=24\left(ans\right)$

$s=12;r_1=4=\frac{\Delta }{s-a}⟹4\times (12-a)=24⟹a=6\left(ans\right)$

$R=\frac{abc}{4\Delta }$

$\tan \frac{A}{2}=\frac{\Delta }{s(s-a)}=\frac{24}{12\times 6}=\frac{1}{3}$

$⟹\frac{A}{2}={\tan }^{-1}\left(\frac{1}{3}\right)⟹A=2{\tan }^{-1}\left(\frac{1}{3}\right)$

$⟹A={\tan }^{-1}\left(\frac{\frac{1}{3}+\frac{1}{3}}{1-\frac{3}{9}}\right)={tan}^{-1}\left(\frac{\frac{2}{3}}{\frac{8}{9}}\right)={tan}^{-1}\left(\frac{2\times 3}{3\times 8}\right)$

$⟹A={\tan }^{-1}\left(\frac{3}{4}\right)={\sin }^{-1}\left(\frac{3}{5}\right)\left(ans\right)$

$2s=a+b+c$

$⟹24=b+b+c⟹b+c=18$

$\Delta =\sqrt{s(s-a)(s-b)(s-c)}=24$

$⟹12\times 6\times (12-b)(12-c)=24\times 24$

$⟹144-12(b+c)+bc=8$

$⟹144-12\times 18+bc=8$

$⟹144-216+bc=8⟹bc-72=8$

$⟹bc=80$

Now;

$R=\frac{abc}{4\Delta }=\frac{6\times 80}{4\times 24}$

$⟹R=5$

$b+\frac{80}{b}=18⟹b^2+80-186=0⟹b^2-186+80=0$

$⟹b=\frac{18\pm \sqrt{324-320}}{2}=\frac{18\pm \sqrt{4}}{2}=\frac{18\pm 2}{2}$

$⟹b=10\text{or}8$

$\therefore c=8\text{or}10$

Again;

$\Delta =\frac{1}{2}ac\times \sin B;[\because \text{taking}c=8]$

$⟹24=\frac{1}{2}\times 6\times 8\times \sin B$

$⟹1=\sin B⟹B=90°$

$\Delta =24=\frac{1}{2}\times 6\times 10\times \sin B$ [Taking e=10]

$⟹B={\sin }^{-1}\frac{4}{5}\left(ans\right)$