Triangle Solving – Problem 3: Multiple choice (select all that apply)

Multiple choice (select all that apply). In $\triangle ABC$, let the sides opposite $A,B,C$ be $a,b,c$. Given $\sin A:\sin B:\sin C=2:3:4$, which statements are true?

A. $(a+b):(b+c):(c+a)=5:6:7$ B. $\triangle ABC$ is obtuse C. If the circumradius is $R$ and the inradius is $r$, then $5R=16r$ D. If $a+b+c=18$, then the area of $\triangle ABC$ is $6\sqrt{15}$

Step 1. Since $\sin A:\sin B:\sin C=2:3:4$, the Law of Sines implies $a:b:c=2:3:4$.

Step 2. Let $a=2t,\ b=3t,\ c=4t$ with $t>0$.

Step 3. Option A: $(a+b):(b+c):(c+a)=5t:7t:6t=5:7:6$, so A is false.

Step 4. Option B: $\cos C=\frac{a^{2}+b^{2}-c^{2}}{2ab}=\frac{4t^{2}+9t^{2}-16t^{2}}{2\cdot2t\cdot3t}=-\frac{1}{4}<0$, so $C$ is obtuse and B is true.

Step 5. Option C: From Step 4, $\sin C=\sqrt{1-\cos^{2}C}=\frac{\sqrt{15}}{4}$. By the Law of Sines, $2R=\frac{c}{\sin C}=\frac{4t}{\sqrt{15}/4}=\frac{16t}{\sqrt{15}}$, so $R=\frac{8t}{\sqrt{15}}$.

Step 6. Using $S=sr$ with $S=\frac12 ab\sin C$ and $s=\frac{a+b+c}{2}$, we get $r=\frac{ab\sin C}{a+b+c}=\frac{(2t)(3t)(\sqrt{15}/4)}{9t}=\frac{\sqrt{15}}{6}t$. Hence $5R=16r$, so C is true.

Step 7. Option D: If $a+b+c=9t=18$, then $t=2$ and $a=4,\ b=6$. The area is $S=\frac12 ab\sin C=\frac12\cdot4\cdot6\cdot\frac{\sqrt{15}}{4}=3\sqrt{15}$, so D is false.

Step 8. Therefore, the correct choices are B and C.

BC