In a triangle ` PQR angle R=(pi)/(4),` if `tan (p/3) and tan ((Q)/(3))` are the roots of the equation `ax ^2 + bx +c=0` then

Options

A. a+b=c
B. b+c=a
C. a+c=b
D. b=c

Answer

To solve the problem step by step, we will analyze the given information and apply the properties of triangles and quadratic equations.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- We have a triangle \( PQR \) with \( \angle R = \frac{\pi}{4} \).
- The roots of the quadratic equation \( ax^2 + bx + c = 0 \) are \( \tan\left(\frac{P}{3}\right) \) and \( \tan\left(\frac{Q}{3}\right) \).

2. **Use the Sum of Angles in a Triangle:**
- In triangle \( PQR \), the sum of the angles is given by:
\[ \angle P + \angle Q + \angle R = \pi \]
- Substituting \( \angle R = \frac{\pi}{4} \):
\[ \angle P + \angle Q + \frac{\pi}{4} = \pi \]
- Rearranging gives:
\[ \angle P + \angle Q = \pi - \frac{\pi}{4} = \frac{3\pi}{4} \]

3. **Divide by 3:**
- Dividing the equation by 3:
\[ \frac{\angle P}{3} + \frac{\angle Q}{3} = \frac{3\pi}{4 \cdot 3} = \frac{\pi}{4} \]

4. **Apply the Tangent Function:**
- Taking the tangent of both sides:
\[ \tan\left(\frac{\angle P}{3} + \frac{\angle Q}{3}\right) = \tan\left(\frac{\pi}{4}\right) \]
- Since \( \tan\left(\frac{\pi}{4}\right) = 1 \), we have:
\[ \tan\left(\frac{\angle P}{3}\right) + \tan\left(\frac{\angle Q}{3}\right) = 1 \]

5. **Use the Sum of Tangents Formula:**
- Using the formula for the sum of tangents:
\[ \tan\left(\frac{P}{3} + \frac{Q}{3}\right) = \frac{\tan\left(\frac{P}{3}\right) + \tan\left(\frac{Q}{3}\right)}{1 - \tan\left(\frac{P}{3}\right) \tan\left(\frac{Q}{3}\right)} \]
- Setting this equal to 1 gives:
\[ \frac{\tan\left(\frac{P}{3}\right) + \tan\left(\frac{Q}{3}\right)}{1 - \tan\left(\frac{P}{3}\right) \tan\left(\frac{Q}{3}\right)} = 1 \]

6. **Cross-Multiply and Rearrange:**
- Cross-multiplying results in:
\[ \tan\left(\frac{P}{3}\right) + \tan\left(\frac{Q}{3}\right) = 1 - \tan\left(\frac{P}{3}\right) \tan\left(\frac{Q}{3}\right) \]
- We know from Vieta's formulas that:
\[ \tan\left(\frac{P}{3}\right) + \tan\left(\frac{Q}{3}\right) = -\frac{b}{a} \]
\[ \tan\left(\frac{P}{3}\right) \tan\left(\frac{Q}{3}\right) = \frac{c}{a} \]

7. **Substituting Vieta's Formulas:**
- Substituting these into our equation gives:
\[ -\frac{b}{a} = 1 - \frac{c}{a} \]
- Multiplying through by \( a \) (assuming \( a \neq 0 \)):
\[ -b = a - c \]
- Rearranging gives:
\[ a + b = c \]

8. **Conclusion:**
- The relationship we derived is \( A + B = C \).

### Final Answer:
Thus, the correct condition is \( A + B = C \).