In a triangle ABC . if $r_1=2 r_2=3 r_3$, then $\sin \mathrm{A}: \sin \mathrm{B}: \sin \mathrm{C}=$

In a triangle ABC . if $r_1=2 r_2=3 r_3$, then $\sin \mathrm{A}: \sin \mathrm{B}: \sin \mathrm{C}=$
  1. $5: 4: 2$
  2. $3: 4: 2$
  3. $6: 3: 2$
  4. $5: 4: 3$

Solution

Given, in $\triangle A B C, r_1=2 r_2=3 r_3$ $\begin{aligned} & \Rightarrow \frac{\Delta}{s-a}=\frac{2 \Delta}{s-b}=\frac{3 \Delta}{s-c}=\frac{1}{K} \text { (Let) } \\ & \Rightarrow \frac{1}{s-a}=\frac{2}{s-b}=\frac{3}{s-c}=\frac{1}{K} \\ & \Rightarrow s-a=K, s-b=2 K, s-c=3 K \\ & \text { and, } s-a+s-b+s-c=6 K \Rightarrow s=6 K \end{aligned}$ So, $6 K-a=K \Rightarrow a=5 K, 6 K-b=2 K \Rightarrow b=4 K$ $6 K-c=3 K \Rightarrow c=3 K$ Since, $\sin A: \sin B: \sin C=\frac{2 \Delta}{c a}: \frac{2 \Delta}{a c}: \frac{2 \Delta}{a b}=5: 4: 3$.

Asked in: AP EAMCET 2024 (19 May Shift 2)

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