Question

In a ABC, if $\frac{s-a}{11}=\frac{s-b}{12}=\frac{s-c}{13}$ then $ta{n}^2\frac{A}{2}$is equal to

• A. $\frac{143}{342}$
• B. $\frac{13}{33}$
• C. $\frac{11}{39}$
• D. $\frac{12}{37}$

Answer 1

Answer: (B)

$\frac{13}{33}$

Let $\frac{s-a}{11}=\frac{s-b}{12}=\frac{s-c}{13}=p\left(Let\right)$

so, $\frac{s-a}{11}=p\Rightarrow a=s-11p..............\left(1\right)$

Similarly, $b=s-12p................\left(2\right)c=s-13p................\left(3\right)$

And we know that $a+b+c=2s$

$\Rightarrow s-11p+s-12p+s-13p=2s\Rightarrow s=36p$

From $\left(1\right),\left(2\right),\left(3\right)s-a=11ps-b=12ps-c=13p$

Also, we know that ,

$ta{n}^2\frac{A}{2}=\frac{(s-b)(s-c)}{\left(s\right)(s-a)}=\frac{12p\times 13p}{36p\times 11p}=\frac{13}{33}$

Therefore, Correct Answer is $B$.