Let $T$ be the tension in the cord and $a$ be the magnitude of the acceleration of the system. We define the positive direction as $m_1$ accelerating up its incline and $m_2$ accelerating down its incline. Applying Newton's second law to each mass parallel to its incline gives two equations:

For mass $m_1$:

$$T - m_1 g \sin(\theta_1) = m_1 a \quad (1)$$

For mass $m_2$:

$$m_2 g \sin(\theta_2) - T = m_2 a \quad (2)$$

We can solve this system for $T$ by first isolating $a$ in each equation: From (1): $a = \frac{T}{m_1} - g \sin(\theta_1)$ From (2): $a = g \sin(\theta_2) - \frac{T}{m_2}$

Setting the expressions for $a$ equal to each other eliminates $a$:

$$\frac{T}{m_1} - g \sin(\theta_1) = g \sin(\theta_2) - \frac{T}{m_2}$$

Now, we solve for $T$:

$$T \left(\frac{1}{m_1} + \frac{1}{m_2}\right) = g (\sin(\theta_1) + \sin(\theta_2))$$ $$T \left(\frac{m_2 + m_1}{m_1 m_2}\right) = g (\sin(\theta_1) + \sin(\theta_2))$$ $$T = \frac{m_1 m_2 g (\sin(\theta_1) + \sin(\theta_2))}{m_1 + m_2}$$