Question

If $R: r_{1}: r=5: 12: 2$, then $r+r_{3}+r_{2}-r_{1}=$

• A. $\cos A$
• B. $\sin A$
• C. $2 r r_{1}$
• D. $2 r_{1}^{2} r$

Answer 1

Answer: (A)

We have, $R: r_{1}: r: 5: 12: 2$
$\therefore R=5 k, r_{1}=12 k, r=2 k$
If $\triangle A B C$ is right angle of $A$
and $a=10 k, b=6 k, c=8 k$
Then $R=5 k, r_{1}=12 k$
and $r=2 k$
Possible
$\therefore a^{2}=b^{2}+c^{2}$
$\therefore r+r_{3}+r_{2}-r_{1}$

$\frac{\Delta}{s}+\frac{\Delta}{s-c}+\frac{\Delta}{s-b}+\frac{\Delta}{s-a}$
$\left(\frac{\Delta}{s}-\frac{\Delta}{s-a}\right)+\left(\frac{\Delta}{s-c}+\frac{\Delta}{s-b}\right)$
$\Rightarrow \Delta\left(\frac{s-a-s}{s(s-a)}+\frac{s-b+s-c}{(s-c)(s-b)}\right)$
$\Rightarrow \Delta\left[\frac{-a}{s(s-a)}+\frac{2 s-b-c}{(s-c)(s-b)}\right]$
$\Rightarrow \Delta\left[\frac{-a}{s(s-a)}+\frac{a}{(s-c)(s-b)}\right]$
$\Rightarrow \Delta a\left[\frac{-(s-c)(s-b)+s(s-a)}{s(s-a)(s-b)(s-c)}\right]$
$\Rightarrow \Delta a \frac{\left(s^{2}-a s-s^{2}+(b+c) s-b c\right)}{\Delta^{2}}$
$\Rightarrow \frac{a^{2}-\left(b^{2}+c^{2}\right)}{\Delta^{2}}=0=\cos A$
$\therefore \cos A$