Question

If $R:r_1:r=5:12:2$, then $r+r_3+r_2-r_1=$

• A. $\cos A$
• B. $\sin A$
• C. $2r{r}_1$
• D. $2r_1^{2}r$

Answer 1

Answer: (A)

We have, $R:r_1:r:5:12:2$
$\therefore R=5k,r_1=12k,r=2k$
If $△ABC$ is right angle of $A$
and $a=10k,b=6k,c=8k$
Then $R=5k,r_1=12k$
and $r=2k$
Possible
$\therefore a^2=b^2+c^2$
$\therefore r+r_3+r_2-r_1$

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