Q68 / 160 · TS EAMCET 2020 (10 Sep Shift 2) · Mathematics

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

  • Correct answer
    A
    $\cos A$
  • B
    $\sin A$
  • C
    $2 r r_1$
  • D
    $2 r_1^2 r$

Solution

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 $\therefore \quad a^2=b^2+c^2$ $\therefore \quad r+r_3+r_2-r_1$ Figure
$\frac{\Delta}{s}+\frac{\Delta}{s-c}+\frac{\Delta}{s-b}+\frac{\Delta}{s-a}$ $\left(\frac{\Delta}{s}-\frac{\Delta}{s-a}\right)+\left(\frac{\Delta}{s-c}+\frac{\Delta}{s-b}\right)$ $\Rightarrow \Delta\left(\frac{s-a-s}{s(s-a)}+\frac{s-b+s-c}{(s-c)(s-b)}\right)$ $\Rightarrow \Delta\left[\frac{-a}{s(s-a)}+\frac{2 s-b-c}{(s-c)(s-b)}\right]$ $\Rightarrow \Delta\left[\frac{-a}{s(s-a)}+\frac{a}{(s-c)(s-b)}\right]$ $\Rightarrow \quad \Delta a\left[\frac{-(s-c)(s-b)+s(s-a)}{s(s-a)(s-b)(s-c)}\right]$ $\Rightarrow \Delta a \frac{\left(s^2-a s-s^2+(b+c) s-b c\right)}{\Delta^2}$ $\Rightarrow \frac{a^2-\left(b^2+c^2\right)}{\Delta^2}=0=\cos A$ $\therefore \cos A$

Hence, the correct answer is option: 'A'.

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