To solve this problem:

We analyze the equilibrium of the block and tensions $T_1$ and $T_2$ based on the diagram.

Step 1: Analyze forces

• The block is in equilibrium, so the vertical and horizontal forces must balance.

Vertical Forces:

• The vertical component of $T_1$ balances the weight $W$ of the block: $T_1 \sin \alpha = W$

Horizontal Forces:

• The horizontal component of $T_1$ balances $T_2$: $T_1 \cos \alpha = T_2$

Step 2: Express $T_1$ in terms of $W$:

From the vertical forces: $T_1 = \frac{W}{\sin \alpha}$

Step 3: Solve for $T_2$:

Using the horizontal forces: $T_2 = T_1 \cos \alpha$ Substitute $T_1 = \frac{W}{\sin \alpha}$: $T_2 = \frac{W \cos \alpha}{\sin \alpha}$

This simplifies to: $T_2 = W \cot \alpha$

Step 4: Compute $T_2$ for $\alpha = 80^\circ$:

The cotangent of $80^\circ$ is approximately $\cot 80^\circ = 0.176$. Thus: $T_2 = 0.176W$

Answer:

The correct choice is: $\boxed{A. \ 0.176W}$

Would you like a detailed breakdown of the calculations or related concepts?

Here are 5 related questions:

1. How would the solution change if $\alpha$ were $45^\circ$?
2. What happens to $T_2$ as $\alpha \to 90^\circ$?
3. Can this setup be used to measure weight $W$ using $T_2$ if $\alpha$ is known?
4. How does the tension $T_1$ compare to $W$ when $\alpha$ increases?
5. What are the real-world assumptions for "massless strings" in equilibrium problems?

Tip: Always check trigonometric values and relationships carefully, especially for equilibrium problems.