Question

If in triangle $ABC,{\Delta }^2=a^2-(b-c{)}^2$ then Let the value of $\tan A$ be $\frac{k}{m}$.Find $m-k$ ?

Answer 1

${\Delta }^2=(a+b-c)(a-b+c)$

$\Rightarrow {\Delta }^2=\left[2\right(s-b\left)2\right(s-c){]}^2$

or $s(s-a)(s-b)(s-c)=16(s-b{)}^2(s-c{)}^2$

or $\frac{(s-b)(s-c)}{s(s-a)}=\frac{1}{16}$

or $\tan \frac{A}{2}=\frac{1}{4}$

$\Rightarrow \tan A=\frac{2\tan \left(\frac{A}{2}\right)}{1-{\tan }^2\left(\frac{A}{2}\right)}=\frac{2\left(\frac{1}{4}\right)}{1-\left(\frac{1}{16}\right)}=\frac{8}{15}$

by comparing it with $\frac{k}{m}$ we get $k=8,m=15$

$\therefore m-k=15-8=7$