Teorema de la tangente

Aplicando el teorema del coseno, podemos calcular el valor de c:

$$\begin{gathered} c^2={12}^2+8^2-2·12·8·\cos 60º \Rightarrow \\ c=\sqrt{144+64-96} \Rightarrow \\ \boxed{c\cong 10.58 cm} \end{gathered}$$

Aplicando el teorema de la tangente, sabemos que:

$$\frac{a+b}{a-b}=\frac{tg {\displaystyle \frac{A+B}{2}}}{tg {\displaystyle \frac{A-B}{2}}}$$

Despejando, obtenemos que:

$$\begin{gathered} \frac{a+b}{a-b}=\frac{tg {\displaystyle \frac{A+B}{2}}}{tg {\displaystyle \frac{A-B}{2}}} \Rightarrow \\ tg \frac{A-B}{2}=\frac{tg \frac{A-B}{2}·(a+b)}{a-b} \end{gathered}$$

Sabiendo que:

$$\begin{gathered} \hat{A}+\hat{B}=180-\hat{C} \Rightarrow \hat{A}+\hat{B} = 120º \Rightarrow \\ \underline{tg \frac{\hat{A}+\hat{B}}{2}=tg 60º} \end{gathered}$$

Entonces:

$$\begin{gathered} \frac{a+b}{a-b}=\frac{tg {\displaystyle \frac{A+B}{2}}}{tg {\displaystyle \frac{A-B}{2}}} \Rightarrow \\ tg \frac{A-B}{2}=\frac{tg 60º·(8+12)}{8-12}\Rightarrow \\ \boxed{tg \frac{A-B}{2}=-5\sqrt{3}} \end{gathered}$$