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In a Delta ABC, angle bisector of exteri...

In a `Delta ABC`, angle bisector of exterior `angle B & angle C` meets at point P, `angle BPC=65^(@)` then find `angleA` .

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To solve for angle A in triangle ABC given the conditions of the problem, we can follow these steps: ### Step-by-Step Solution: 1. **Understanding the Problem**: We have triangle ABC where the angle bisectors of the exterior angles at B and C meet at point P. We know that angle BPC = 65°. 2. **Using the Property of Exterior Angles**: The angle bisector of an exterior angle divides it into two equal parts. Therefore, if we denote the exterior angle at B as \( \angle B_{ext} \) and at C as \( \angle C_{ext} \), we can write: - \( \angle B_{ext} = 180° - \angle B \) - \( \angle C_{ext} = 180° - \angle C \) 3. **Setting Up the Angles**: Since P is the intersection of the angle bisectors of the exterior angles, we can express: - \( \angle BPC = \frac{1}{2} \left( \angle B_{ext} + \angle C_{ext} \right) \) 4. **Substituting the Values**: We know that: \[ \angle BPC = 65° \] Therefore, we can set up the equation: \[ 65° = \frac{1}{2} \left( (180° - \angle B) + (180° - \angle C) \right) \] 5. **Simplifying the Equation**: This simplifies to: \[ 65° = \frac{1}{2} \left( 360° - (\angle B + \angle C) \right) \] Multiplying both sides by 2 gives: \[ 130° = 360° - (\angle B + \angle C) \] Rearranging gives: \[ \angle B + \angle C = 360° - 130° = 230° \] 6. **Using the Triangle Sum Property**: In triangle ABC, the sum of the angles is: \[ \angle A + \angle B + \angle C = 180° \] Substituting \( \angle B + \angle C = 230° \) into this equation gives: \[ \angle A + 230° = 180° \] 7. **Solving for Angle A**: Rearranging gives: \[ \angle A = 180° - 230° = -50° \] This result is not possible, indicating a mistake in the interpretation of angles. 8. **Correcting the Approach**: Instead, we should use the relationship: \[ 90° - \frac{\angle A}{2} = 65° \] Rearranging gives: \[ \frac{\angle A}{2} = 90° - 65° = 25° \] Thus: \[ \angle A = 25° \times 2 = 50° \] ### Final Answer: \[ \angle A = 50° \]
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