In triangle `ABC, /_ABC = 90^@, BP ` is drawn perpendicular to AC. If `/_BAP = 50^@`, what is the value (in degree) of `/_PBC`?

Options

A. `30`
B. `45`
C. `50 `
D. `60 `

Answer

To solve the problem, we will follow these steps:

1. **Understand the Triangle Configuration**: We have triangle ABC with angle ABC = 90°. BP is drawn perpendicular to AC, creating point P on AC. We know that angle BAP = 50°.

2. **Identify the Angles**: In triangle ABC, we have:
- Angle ABC = 90°
- Angle BAP = 50°
- Therefore, angle BAC can be calculated as follows:
\[ \text{Angle BAC} = 90° - \text{Angle BAP} = 90° - 50° = 40° \]

3. **Consider Triangle PBC**: Now, we need to find angle PBC. In triangle PBC, we have:
- Angle PBC (which we need to find) + Angle BPC + Angle BCP = 180°

4. **Identify Angle BPC**: Since BP is perpendicular to AC, angle BPC = 90°.

5. **Identify Angle BCP**: We already calculated that angle BAC = 40°. Since angle BAP = 50°, angle BCP can be calculated as:
\[ \text{Angle BCP} = \text{Angle BAC} = 40° \]

6. **Set Up the Equation**: Now we can set up the equation for triangle PBC:
\[ \text{Angle PBC} + 90° + 40° = 180° \]

7. **Solve for Angle PBC**:
\[ \text{Angle PBC} + 130° = 180° \]
\[ \text{Angle PBC} = 180° - 130° = 50° \]

Thus, the value of angle PBC is **50°**.