Question

In a triangle \(ABC\), if \(\tan\tfrac{A}{2} : \tan\tfrac{B}{2} : \tan\tfrac{C}{2} = 15 : 10 : 6\), then \(\frac{a}{b - c} = \,?\)

• A. \(\tfrac{8}{3}\)
• B. \(\tfrac{7}{3}\)
• C. \(5\)
• D. \(4\)

Hint:

In triangle geometry, \(\tan(\tfrac{A}{2}):\tan(\tfrac{B}{2}):\tan(\tfrac{C}{2})\) often converts to \((s-a)^{-1}:(s-b)^{-1}:(s-c)^{-1}\).
- Remember \(a+b+c=2s\). Then solve consistently for the side lengths.

Answer 1

The Correct Option is D

Step 1: Recall half-angle formula relations.
In a triangle with sides \(a,b,c\) opposite angles \(A,B,C\), we have: \[ \tan\tfrac{A}{2} : \tan\tfrac{B}{2} : \tan\tfrac{C}{2} \;\;=\;\; \frac{r}{s-a} : \frac{r}{s-b} : \frac{r}{s-c}, \] where \(r\) is the inradius and \(s=\tfrac{a+b+c}{2}\) is the semiperimeter. So \[ \tan\tfrac{A}{2} \;=\;\frac{r}{s-a}, \quad \tan\tfrac{B}{2} \;=\;\frac{r}{s-b}, \quad \tan\tfrac{C}{2} \;=\;\frac{r}{s-c}. \] Thus \[ \tan\tfrac{A}{2} : \tan\tfrac{B}{2} : \tan\tfrac{C}{2} = \frac{1}{s-a} : \frac{1}{s-b} : \frac{1}{s-c}. \] 

Step 2: Given ratio and deduce \((s-a),(s-b),(s-c)\).
We have \(\tan\tfrac{A}{2} : \tan\tfrac{B}{2} : \tan\tfrac{C}{2} = 15 : 10 : 6\). This implies \[ \frac{1}{s-a} : \frac{1}{s-b} : \frac{1}{s-c} = 15 : 10 : 6. \] So let \(\;s-a = \tfrac{1}{15}k,\; s-b = \tfrac{1}{10}k,\; s-c = \tfrac{1}{6}k\) for some \(k\). Hence \[ a = s - \tfrac{k}{15},\quad b = s - \tfrac{k}{10},\quad c = s - \tfrac{k}{6}. \] 

Step 3: Also recall \(a+b+c = 2s\).
So \[ (s - \tfrac{k}{15}) + (s - \tfrac{k}{10}) + (s - \tfrac{k}{6}) = 2s. \] This simplifies to \[ 3s - \Bigl(\tfrac{k}{15}+\tfrac{k}{10}+\tfrac{k}{6}\Bigr) = 2s \quad\Longrightarrow\quad s = \tfrac{k}{15} + \tfrac{k}{10} + \tfrac{k}{6}. \] Compute that sum carefully: \[ \frac{k}{15} + \frac{k}{10} + \frac{k}{6} = k\Bigl(\frac{1}{15}+\frac{1}{10}+\frac{1}{6}\Bigr) = k\Bigl(\frac{2}{30}+\frac{3}{30}+\frac{5}{30}\Bigr) = k\cdot \frac{10}{30} = \frac{k}{3}. \] Hence \(s=\tfrac{k}{3}\). Step 4: Express \(a,\,b,\,c\) in terms of \(k\).
\[ a = \frac{k}{3} - \frac{k}{15} = \frac{5k - k}{15} = \frac{4k}{15},\quad b = \frac{k}{3} - \frac{k}{10} = \frac{10k -3k}{30} = \frac{7k}{30}\times 2 = \frac{7k}{30}\cdot 2\ \] Better to do each carefully: - \(a = s - \tfrac{k}{15} = \tfrac{k}{3} - \tfrac{k}{15} = \tfrac{5k}{15} - \tfrac{k}{15} = \tfrac{4k}{15}.\) - \(b = s - \tfrac{k}{10} = \tfrac{k}{3} - \tfrac{k}{10} = \tfrac{10k}{30} - \tfrac{3k}{30} = \tfrac{7k}{30}.\) - \(c = s - \tfrac{k}{6} = \tfrac{k}{3} - \tfrac{k}{6} = \tfrac{2k}{6} - \tfrac{k}{6} = \tfrac{k}{6}.\) So \[ a = \frac{4k}{15},\quad b = \frac{7k}{30},\quad c = \frac{k}{6}. \] (We can also double-check \(b\) in simpler fraction form if desired.) 

Step 5: Find \(\tfrac{a}{b-c}\).
\[ b-c = \frac{7k}{30} - \frac{k}{6} = \frac{7k}{30} - \frac{5k}{30} = \frac{2k}{30} = \frac{k}{15}. \] Hence \[ \frac{a}{b-c} = \frac{\tfrac{4k}{15}}{\tfrac{k}{15}} = 4. \] Thus \(\boxed{4}\) is the value of \(\tfrac{a}{b-c}\).