In the sport of parasailing, a person is attached to two ropes: one rope attaches to the boat by which the parasailor is being pulled, and the other rope attaches to a parachute-like sail. Each of the two ropes is coming off the parasailor at a different angle.
 
A rider is towed at a constant speed by a rope attached to the boat which makes an angle of 12º with the horizon. The tension in this rope is 1,068 N. At the same time, the rope to the sail makes an angle of 34º with the horizon. What is the weight (not the mass) of the parasailor?

To determine the weight of the parasailor, we need to analyze the forces acting on the parasailor and apply some principles of vector mechanics. There are three main forces to consider:

1. The **tension** in the rope from the boat (\(T_B = 1068 \, \text{N}\)), which makes an angle of \(12^\circ\) with the horizontal.
2. The **tension** in the rope from the parachute-like sail (\(T_S\)), which makes an angle of \(34^\circ\) with the horizontal.
3. The **weight** of the parasailor (\(W\)), acting vertically downward due to gravity.

Since the parasailor is being towed at a constant speed, the forces in both the horizontal and vertical directions must balance. Therefore, we can break this problem down into horizontal (\(x\)) and vertical (\(y\)) components of the forces and use equilibrium conditions to find the weight of the parasailor.

### Step 1: Break the forces into components
#### Horizontal components
- The horizontal component of the force from the boat: 
  \[
  T_B^x = T_B \cos(12^\circ)
  \]
- The horizontal component of the force from the sail: 
  \[
  T_S^x = T_S \cos(34^\circ)
  \]
Since the parasailor is moving at a constant speed, the net horizontal force is zero, so the horizontal forces must balance:
\[
T_B^x = T_S^x
\]
This gives:
\[
T_B \cos(12^\circ) = T_S \cos(34^\circ)
\]
Substitute \(T_B = 1068 \, \text{N}\):
\[
1068 \cos(12^\circ) = T_S \cos(34^\circ)
\]
Solving for \(T_S\):
\[
T_S = \frac{1068 \cos(12^\circ)}{\cos(34^\circ)}
\]
\[
T_S \approx \frac{1068 \times 0.9781}{0.8290} \approx 1259.4 \, \text{N}
\]

### Step 2: Vertical force equilibrium
Now, in the vertical direction, the forces must also balance. The upward forces are the vertical components of the tension in the ropes from the boat and the sail, while the downward force is the weight of the parasailor.

The vertical components of the forces are:
- From the boat:
  \[
  T_B^y = T_B \sin(12^\circ)
  \]
- From the sail:
  \[
  T_S^y = T_S \sin(34^\circ)
  \]
- The weight of the parasailor acts downward: \(W\)

The net vertical force is zero, so the vertical forces must balance:
\[
T_B^y + T_S^y = W
\]
Substituting the values for \(T_B\) and \(T_S\):
\[
1068 \sin(12^\circ) + 1259.4 \sin(34^\circ) = W
\]
Calculate the vertical components:
\[
1068 \sin(12^\circ) \approx 1068 \times 0.2079 = 221.9 \, \text{N}
\]
\[
1259.4 \sin(34^\circ) \approx 1259.4 \times 0.5592 = 703.9 \, \text{N}
\]
Now sum these components to find the weight:
\[
W = 221.9 + 703.9 = 925.8 \, \text{N}
\]

Thus, the weight of the parasailor is approximately \(925.8 \, \text{N}\).

### Summary:
The weight of the parasailor is \(W \approx 925.8 \, \text{N}\).

Would you like a breakdown of the steps or have any questions on the approach?

Here are some related questions you could explore:
1. How do you calculate the mass of the parasailor given the weight?
2. What happens if the angle of the boat rope increases while keeping the same tension?
3. How would the force balance change if the parasailor accelerates?
4. What is the importance of constant speed in this problem?
5. How would adding wind resistance affect the force balance?

**Tip:** When solving force equilibrium problems, breaking the forces into components along common directions (horizontal and vertical) simplifies the analysis.

In the sport of parasailing, a person is attached to two ropes: one rope attaches to the boat by which the parasailor is being pulled, and the other rope attaches to a parachute-like sail. Each of the two ropes is coming off the parasailor at a different angle.

A rider is towed at a constant speed by a rope attached to the boat which makes an angle of 12º with the horizon. The tension in this rope is 1,068 N. At the same time, the rope to the sail makes an angle of 34º with the horizon. What is the weight (not the mass) of the parasailor?

Learn how to calculate the weight of a parasailor using vector decomposition and force equilibrium. This example demonstrates how to apply trigonometry to balance the forces in parasailing, using angles of 12º and 34º, and rope tensions. Suitable for high school students in Grades 10-12.

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