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To solve for the values of \( P \) and \( Q \) in this equilibrium problem, we need to apply the conditions for equilibrium. The forces are balanced, meaning both the horizontal and vertical components of the forces must sum to zero.

### Step 1: Resolve the forces into components

The given forces are:
- A force of 10 N at an angle of 49° above the horizontal.
- Forces \( P \) and \( Q \), whose directions are given as vertical and angled 18° below the horizontal, respectively.

#### Force Components:
- **10 N Force**:
  - Horizontal component: \( 10 \cos(49^\circ) \)
  - Vertical component: \( 10 \sin(49^\circ) \)

- **Force \( Q \)** (angled at 18° below the horizontal):
  - Horizontal component: \( Q \cos(18^\circ) \)
  - Vertical component: \( Q \sin(18^\circ) \)

- **Force \( P \)** is purely vertical, so it only has a vertical component \( P \).

### Step 2: Set up equilibrium conditions

#### Horizontal equilibrium:
Since the system is in equilibrium, the sum of horizontal forces must be zero:
\[
10 \cos(49^\circ) = Q \cos(18^\circ)
\]
From this, we can solve for \( Q \):
\[
Q = \frac{10 \cos(49^\circ)}{\cos(18^\circ)}
\]

#### Vertical equilibrium:
The sum of the vertical forces must also be zero:
\[
P + Q \sin(18^\circ) = 10 \sin(49^\circ)
\]
We can rearrange this to solve for \( P \):
\[
P = 10 \sin(49^\circ) - Q \sin(18^\circ)
\]

### Step 3: Calculation

Using the known trigonometric values:
- \( \cos(49^\circ) \approx 0.6561 \)
- \( \sin(49^\circ) \approx 0.7547 \)
- \( \cos(18^\circ) \approx 0.9511 \)
- \( \sin(18^\circ) \approx 0.3090 \)

#### Solving for \( Q \):
\[
Q = \frac{10 \times 0.6561}{0.9511} \approx 6.90 \, \text{N}
\]

#### Solving for \( P \):
\[
P = 10 \times 0.7547 - 6.90 \times 0.3090 \approx 7.55 - 2.13 = 5.42 \, \text{N}
\]

### Final Answer:
- \( Q = 6.9 \, \text{N} \)
- \( P = 5.4 \, \text{N} \)

Would you like a detailed explanation of the steps or any clarifications?

Here are 5 follow-up questions:
1. What happens if the angle of force \( Q \) is changed?
2. How do we solve for forces in non-equilibrium cases?
3. Can this method be used for three-dimensional force problems?
4. How do the trigonometric identities simplify force resolution?
5. What happens if one of the forces is doubled?

**Tip:** Always break forces into their components when dealing with equilibrium problems involving angled forces.

Find the values of P and Q from the given forces in equilibrium with angles 49° and 18°.

Solve for the values of P and Q in a system of forces in equilibrium. This problem demonstrates how to apply trigonometric methods to resolve forces and find equilibrium conditions. Suitable for students in Grades 10-12.

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