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Assertion (A): In $\triangle A B C$, if $r=6, r_2=36, R=15$ then $c^2+a^2=b^2$
Reason (R): In $\triangle A B C$, if $r: R: r_2=1: 2.5: 6$ then $\mathrm{B}=90^{\circ}$
The correct option among the following is
Options:
Reason (R): In $\triangle A B C$, if $r: R: r_2=1: 2.5: 6$ then $\mathrm{B}=90^{\circ}$
The correct option among the following is
Solution:
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Verified Answer
The correct answer is:
Both (A) and (R) are true. (R) is a correct explanation of (A)
(R) Given that $\mathrm{r}: \mathrm{R}: \mathrm{r}_2=1: 2.5: 6=2: 5: 12$
$\Rightarrow \mathrm{r}=2 \mathrm{k}, \mathrm{R}=5 \mathrm{k}$ and $\mathrm{r}_2=12 \mathrm{k}$
we have $r_2-r=4 R \sin ^2 \frac{B}{2}$
$\begin{aligned} & \Rightarrow \quad 12 \mathrm{k}-2 \mathrm{k}=4 \times 5 \mathrm{k} \sin ^2 \frac{\mathrm{B}}{2} \\ & \Rightarrow \quad 10 \mathrm{k}=20 \mathrm{k} \sin ^2 \frac{\mathrm{B}}{2} \Rightarrow \sin \frac{\mathrm{B}}{2}=\frac{1}{\sqrt{2}} \Rightarrow 90^{\circ}\end{aligned}$
$\therefore \quad$ Reason is true
Hence assertion is also true.
$\Rightarrow \mathrm{r}=2 \mathrm{k}, \mathrm{R}=5 \mathrm{k}$ and $\mathrm{r}_2=12 \mathrm{k}$
we have $r_2-r=4 R \sin ^2 \frac{B}{2}$
$\begin{aligned} & \Rightarrow \quad 12 \mathrm{k}-2 \mathrm{k}=4 \times 5 \mathrm{k} \sin ^2 \frac{\mathrm{B}}{2} \\ & \Rightarrow \quad 10 \mathrm{k}=20 \mathrm{k} \sin ^2 \frac{\mathrm{B}}{2} \Rightarrow \sin \frac{\mathrm{B}}{2}=\frac{1}{\sqrt{2}} \Rightarrow 90^{\circ}\end{aligned}$
$\therefore \quad$ Reason is true
Hence assertion is also true.
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