Question

In a triangle \(ABC\), if \[ \tan\frac{A}{2}:\tan\frac{B}{2}:\tan\frac{C}{2}=1:2:3, \] then \[ \frac{a+3c}{b}= \]

• A. \(4\)
• B. \(3\)
• C. \(2\)
• D. \(6\)

Hint:

For triangle problems involving \(\tan\frac A2,\tan\frac B2,\tan\frac C2\), remember: \[ \tan\frac A2:\tan\frac B2:\tan\frac C2 = \frac1{s-a}:\frac1{s-b}:\frac1{s-c}. \] This converts trigonometric ratios directly into side relations.

Answer 1

The Correct Option is A

Concept: In any triangle, \[ \tan\frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}, \] \[ \tan\frac{B}{2}=\sqrt{\frac{(s-c)(s-a)}{s(s-b)}}, \] \[ \tan\frac{C}{2}=\sqrt{\frac{(s-a)(s-b)}{s(s-c)}}. \] A very useful consequence is \[ \tan\frac{A}{2}:\tan\frac{B}{2}:\tan\frac{C}{2} = \frac1{s-a}:\frac1{s-b}:\frac1{s-c}. \] This relation allows us to convert the given ratio into relations among the sides.

Step 1: Use the given ratio. Given, \[ \tan\frac{A}{2}:\tan\frac{B}{2}:\tan\frac{C}{2} = 1:2:3. \] Hence, \[ \frac1{s-a}:\frac1{s-b}:\frac1{s-c} = 1:2:3. \] Therefore, \[ (s-a):(s-b):(s-c) = 6:3:2. \] Let \[ s-a=6k,\qquad s-b=3k,\qquad s-c=2k. \]

Step 2: Express sides in terms of \(k\). Adding, \[ (s-a)+(s-b)+(s-c)=11k. \] But \[ (s-a)+(s-b)+(s-c)=s. \] Hence \[ s=11k. \] Therefore, \[ a=s-6k=5k, \] \[ b=s-3k=8k, \] \[ c=s-2k=9k. \]

Step 3: Evaluate the required expression. \[ \frac{a+3c}{b} = \frac{5k+3(9k)}{8k} = \frac{32k}{8k} = 4. \] Conclusion: \[ \boxed{4} \]