Explain in English

For the 8 kg mass on the left incline:
The component of weight acting down the incline is $W_1 = 8g \sin\theta$.
The tension in the left string, $T_1$, acts up the incline.
Since the mass is in equilibrium, $T_1 = 8g \sin\theta$.

For the 6 kg mass on the right incline:
The component of weight acting down the incline is $W_2 = 6g \sin\beta$.
The tension in the right string, $T_2$, acts up the incline.
Since the mass is in equilibrium, $T_2 = 6g \sin\beta$.

The 4 kg mass hangs freely, so its weight acts downwards: $W_3 = 4g$.
The left string makes an angle of $60^\circ$ with the upward vertical.
The right string makes an angle of $60^\circ$ with the upward vertical.
For equilibrium, the sum of vertical forces is zero and the sum of horizontal forces is zero.

Vertical equilibrium:

$$T_1 \cos(60^\circ) + T_2 \cos(60^\circ) = 4g$$

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

$$\frac{1}{2}T_1 + \frac{1}{2}T_2 = 4g$$ $$T_1 + T_2 = 8g \quad \text{(Equation 1)}$$

Horizontal equilibrium:

$$T_1 \sin(60^\circ) = T_2 \sin(60^\circ)$$

Since $\sin(60^\circ) \neq 0$, we can divide by it:

$$T_1 = T_2 \quad \text{(Equation 2)}$$

Substitute Equation 2 into Equation 1:

$$T_1 + T_1 = 8g$$ $$2T_1 = 8g$$ $$T_1 = 4g$$

From Equation 2, $T_2 = 4g$.

Using the equilibrium condition for the 8 kg mass:

$$T_1 = 8g \sin\theta$$ $$4g = 8g \sin\theta$$ $$\sin\theta = \frac{4g}{8g} = \frac{1}{2}$$ $$\theta = \arcsin\left(\frac{1}{2}\right) = \boxed{30^\circ}$$

Using the equilibrium condition for the 6 kg mass:

$$T_2 = 6g \sin\beta$$ $$4g = 6g \sin\beta$$ $$\sin\beta = \frac{4g}{6g} = \frac{2}{3}$$ $$\beta = \arcsin\left(\frac{2}{3}\right) \approx \boxed{41.8^\circ}$$

:
The system is in equilibrium. Let $g$ be the acceleration due to gravity.

Both planes have an inclination of $30^\circ$.
For mass A ($m_A = 2m$):
The component of weight acting down the incline is $W_A = m_A g \sin(30^\circ) = (2m)g\left(\frac{1}{2}\right) = mg$.
The tension in the left string, $T_A$, acts up the incline.
Since mass A is in equilibrium, $T_A = mg$.

For mass B ($m_B = m$):
The component of weight acting down the incline is $W_B = m_B g \sin(30^\circ) = (m)g\left(\frac{1}{2}\right) = \frac{1}{2}mg$.
The tension in the right string, $T_B$, acts up the incline.
Since mass B is in equilibrium, $T_B = \frac{1}{2}mg$.

Mass C ($m_C = m$) hangs freely, so its weight acts downwards: $W_C = mg$.
The left string makes an angle $\theta$ with the upward vertical.
The right string makes an angle $\beta$ with the upward vertical.
For equilibrium, the sum of vertical forces is zero and the sum of horizontal forces is zero.

Vertical equilibrium:

$$T_A \cos\theta + T_B \cos\beta = W_C$$

Substitute $T_A = mg$, $T_B = \frac{1}{2}mg$, and $W_C = mg$:

$$mg \cos\theta + \frac{1}{2}mg \cos\beta = mg$$

Divide by $mg$:

$$\cos\theta + \frac{1}{2}\cos\beta = 1$$

Multiply by 2:

$$2\cos\theta + \cos\beta = 2$$

This can be written as $\boxed{\cos\beta + 2\cos\theta = 2}$, which matches the second relation to show.

Horizontal equilibrium:

$$T_A \sin\theta = T_B \sin\beta$$

Substitute $T_A = mg$ and $T_B = \frac{1}{2}mg$:

$$mg \sin\theta = \frac{1}{2}mg \sin\beta$$

Divide by $mg$:

$$\sin\theta = \frac{1}{2}\sin\beta$$

This can be written as $\boxed{2\sin\theta = \sin\beta}$.
Note: The problem statement asks to show $\sin\theta = 2\sin\beta$. There is a discrepancy here. Based on the diagram and given masses, the correct relation is $2\sin\theta = \sin\beta$. We will use this derived relation to find $\theta$ and $\beta$.

We have two equations:

1. $2\sin\theta = \sin\beta$
2. $2\cos\theta + \cos\beta = 2$

From (1), $\sin\beta = 2\sin\theta$.
From (2), $\cos\beta = 2 - 2\cos\theta$.

Using the trigonometric identity $\sin^2\beta + \cos^2\beta = 1$:

$$(2\sin\theta)^2 + (2 - 2\cos\theta)^2 = 1$$ $$4\sin^2\theta + (4 - 8\cos\theta + 4\cos^2\theta) = 1$$ $$4(\sin^2\theta + \cos^2\theta) - 8\cos\theta + 4 = 1$$

Since $\sin^2\theta + \cos^2\theta = 1$:

$$4(1) - 8\cos\theta + 4 = 1$$ $$8 - 8\cos\theta = 1$$ $$8\cos\theta = 7$$ $$\cos\theta = \frac{7}{8}$$ $$\theta = \arccos\left(\frac{7}{8}\right) \approx \boxed{29.0^\circ}$$

Now find $\beta$ using $\cos\beta = 2 - 2\cos\theta$:

$$\cos\beta = 2 - 2\left(\frac{7}{8}\right)$$ $$\cos\beta = 2 - \frac{14}{8}$$ $$\cos\beta = 2 - \frac{7}{4}$$ $$\cos\beta = \frac{8}{4} - \frac{7}{4}$$ $$\cos\beta = \frac{1}{4}$$ $$\beta = \arccos\left(\frac{1}{4}\right) \approx \boxed{75.5^\circ}$$