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To solve for the reaction forces \( R_A \) (vertical reaction at point A), \( R_H \) (horizontal reaction at point A), and \( R_C \) (reaction at point C), we need to apply static equilibrium conditions. This is a typical statics problem where we assume the beam is in equilibrium under the action of the applied force and the reaction forces at the supports.

### Step-by-step process:

#### 1. **Given Data**:
- A horizontal beam of length \( 18 \, \text{m} \) (divided into 10 m from A to B and 8 m from B to C).
- Applied force \( P = 160 \, \text{kN} \) at point B.
- Point A is a hinged support, so it can have both vertical (\( R_A \)) and horizontal (\( R_H \)) reactions.
- Point C is a roller support, so it only has a vertical reaction (\( R_C \)).

The force \( P \) is inclined, and we can break it into horizontal and vertical components:
- \( P_v = 160 \, \text{kN} \cdot \sin(\theta) \)
- \( P_h = 160 \, \text{kN} \cdot \cos(\theta) \)

However, the angle \( \theta \) is not given, so I will assume the angle to be 45 degrees for calculation purposes (unless a specific angle is provided).

Thus,
- \( P_v = P_h = 160 \, \text{kN} \cdot \frac{1}{\sqrt{2}} = 113.14 \, \text{kN} \)

#### 2. **Equilibrium Conditions**:
The beam is in static equilibrium, so we apply the following three conditions:
- **Sum of horizontal forces = 0**:
  \[
  \sum F_x = 0 \implies R_H - P_h = 0
  \]
  So,
  \[
  R_H = P_h = 113.14 \, \text{kN}
  \]

- **Sum of vertical forces = 0**:
  \[
  \sum F_y = 0 \implies R_A + R_C - P_v = 0
  \]
  Substituting \( P_v = 113.14 \, \text{kN} \),
  \[
  R_A + R_C = 113.14 \, \text{kN}
  \]
  (Equation 1)

- **Sum of moments about point A = 0**:
  Taking moments about point A (clockwise moments are positive):
  \[
  \sum M_A = 0 \implies -P_v \times 10 \, \text{m} + R_C \times 18 \, \text{m} = 0
  \]
  Substituting \( P_v = 113.14 \, \text{kN} \),
  \[
  -113.14 \, \text{kN} \times 10 + R_C \times 18 = 0
  \]
  Solving for \( R_C \):
  \[
  R_C = \frac{1131.4}{18} = 62.86 \, \text{kN}
  \]

#### 3. **Solving for \( R_A \)**:
From Equation 1:
\[
R_A + 62.86 = 113.14
\]
\[
R_A = 113.14 - 62.86 = 50.28 \, \text{kN}
\]

### Final Answers:
- \( R_H = 113.14 \, \text{kN} \)
- \( R_A = 50.28 \, \text{kN} \)
- \( R_C = 62.86 \, \text{kN} \)

Do you want further details or clarification on any part of this solution?

Here are five related questions:
1. What happens if the applied force \( P \) is at a different angle?
2. How would the reaction forces change if the length of the beam from A to B or B to C changes?
3. How do we handle situations where the beam is not horizontal but inclined?
4. How can we calculate the internal moments and shear forces at any point on the beam?
5. What if there were multiple loads applied at different points on the beam?

**Tip:** Always start by resolving forces into their horizontal and vertical components to simplify the equilibrium calculations.

Find the reaction forces RA, RH, and RC for the beam with an applied force of 160 kN at point B, and supports at points A and C.

This problem involves finding the reaction forces at supports A and C for a beam with a 160 kN inclined force at point B. Using static equilibrium conditions, we calculate vertical and horizontal components of the force and apply the sum of forces and moments equations.

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