If $A+B+C=180$° and $\tan\frac A2=\frac 13$ and $\tan\frac B2= \frac 23$, then $\tan\frac C2$ is

Question

If $A+B+C=180$° and $\tan\frac A2=\frac 13$ and $\tan\frac B2= \frac 23$, then $\tan\frac C2$ is

My solution

$$\tan2A=\frac{2\tan A}{1-\tan^2A}$$

So $$\tan A=\frac 34$$ and $$\tan B=\frac {12}{5}$$ Since it’s a triangle $$\tan A+\tan B+\tan C=\tan A\tan B\tan C$$ Substituting the values $$\frac 34+ \frac{12}{5}+ \tan C=\frac 95 \tan C$$ We get the value of $\tan C$ as $\frac {53}{16}$

Unsurprisingly enough, the actually answer is $\frac 79$. How do I obtain the right answer?

Answer 1

You have a slight error: $\frac{4}{5} \tan C = \frac{63}{20}$ so $\tan C = \frac{63}{16}$, which leads to $\tan C/2 = \frac{7}{9}$.

However, apart from that all your steps are correct.

You can also verify this answer by direct computation: $\tan \big(1/2(\pi - 2 \arctan 1/3 - 2 \arctan 2/3) \big) = \frac{7}{9}$.

Answer 2

$\tan C$ is going to be $\dfrac{63}{16}$, not $\dfrac{53}{16}$. After that, one can calculate $\tan \frac{C}{2}$ which will be $\dfrac{7}{9}$.