Two masses, 8kg and 6kg, are attached to the ends of a light inextensible string which passes over a smooth pulley. A third mass of 4kg is suspended from the pulley by another string, and this mass is

The 4kg mass is in equilibrium. The weight of the mass acts downwards, and the vertical components of the tension in the two strings act upwards. Each string makes an angle of $60^\circ$ with the upward vertical. Let $T$ be the tension in each string.
The weight of the 4kg mass is $W = m g = 4g$.
For vertical equilibrium:

$$2T \cos(60^\circ) = 4g$$

Since $\cos(60^\circ) = \frac{1}{2}$:

$$2T \left(\frac{1}{2}\right) = 4g$$ $$T = 4g$$

The 8kg mass is in equilibrium on a smooth inclined plane. The tension $T$ pulls the mass up the plane, and the component of gravity pulls it down the plane.
The component of gravity acting down the incline is $m_1 g \sin(\theta) = 8g \sin(\theta)$.
For equilibrium along the incline:

$$T = 8g \sin(\theta)$$

Substitute the value of $T$ from Step 1:

$$4g = 8g \sin(\theta)$$

Divide both sides by $8g$:

$$\sin(\theta) = \frac{4g}{8g} = \frac{1}{2}$$ $$\theta = \arcsin\left(\frac{1}{2}\right)$$ $$\theta = 30^\circ$$

The 6kg mass is in equilibrium on a smooth inclined plane. The tension $T$ pulls the mass up the plane, and the component of gravity pulls it down the plane.
The component of gravity acting down the incline is $m_2 g \sin(\beta) = 6g \sin(\beta)$.
For equilibrium along the incline:

$$T = 6g \sin(\beta)$$

Substitute the value of $T$ from Step 1:

$$4g = 6g \sin(\beta)$$

Divide both sides by $6g$:

$$\sin(\beta) = \frac{4g}{6g} = \frac{2}{3}$$ $$\beta = \arcsin\left(\frac{2}{3}\right)$$

Using a calculator:

$$\beta \approx 41.81^\circ$$

The angles are:

$$\theta = 30^\circ, \beta \approx 41.8^\circ$$