In `Delta PQR` , `/_R=pi/4`, `tan(P/3)`, `tan(Q/3)` are the roots of the equation `ax^2+bx+c=0`, then

Options

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

Answer

To solve the problem, we need to find the relationship between the coefficients \( a \), \( b \), and \( c \) of the quadratic equation \( ax^2 + bx + c = 0 \) given that \( \tan(P/3) \) and \( \tan(Q/3) \) are the roots of the equation, and \( \angle R = \frac{\pi}{4} \).

### Step-by-Step Solution:

1. **Understand the Triangle Properties**:
In triangle \( PQR \), we know that the sum of the angles is \( \pi \):
\[ P + Q + R = \pi \]
Given \( R = \frac{\pi}{4} \), we can express \( P + Q \):
\[ P + Q = \pi - R = \pi - \frac{\pi}{4} = \frac{3\pi}{4} \]

2. **Identify the Roots**:
The roots of the quadratic equation are given as \( \tan\left(\frac{P}{3}\right) \) and \( \tan\left(\frac{Q}{3}\right) \).

3. **Use the Sum and Product of Roots**:
For a quadratic equation \( ax^2 + bx + c = 0 \), the sum and product of the roots can be expressed as:
- Sum of the roots:
\[ \tan\left(\frac{P}{3}\right) + \tan\left(\frac{Q}{3}\right) = -\frac{b}{a} \]
- Product of the roots:
\[ \tan\left(\frac{P}{3}\right) \cdot \tan\left(\frac{Q}{3}\right) = \frac{c}{a} \]

4. **Calculate the Sum of the Roots**:
Using the tangent addition formula:
\[ \tan\left(\frac{P}{3} + \frac{Q}{3}\right) = \tan\left(\frac{P + Q}{3}\right) = \tan\left(\frac{3\pi/4}{3}\right) = \tan\left(\frac{\pi}{4}\right) = 1 \]
Therefore, we have:
\[ \tan\left(\frac{P}{3}\right) + \tan\left(\frac{Q}{3}\right) = 1 \]
Thus,
\[ 1 = -\frac{b}{a} \implies b = -a \]

5. **Calculate the Product of the Roots**:
Using the product of tangents:
\[ \tan\left(\frac{P}{3}\right) \cdot \tan\left(\frac{Q}{3}\right) = \frac{\tan\left(\frac{P + Q}{3}\right)}{1 - \tan\left(\frac{P}{3}\right) \tan\left(\frac{Q}{3}\right)} = \frac{1}{1 - \tan\left(\frac{P}{3}\right) \tan\left(\frac{Q}{3}\right)} \]
Let \( x = \tan\left(\frac{P}{3}\right) \) and \( y = \tan\left(\frac{Q}{3}\right) \). Then:
\[ xy = \frac{c}{a} \]
Since \( P + Q = \frac{3\pi}{4} \), we can find \( \tan\left(\frac{3\pi/4}{3}\right) \) and relate it back to \( c \).

6. **Final Relationship**:
From the previous steps, we have:
\[ c = a + b \]
Since \( b = -a \), we can substitute:
\[ c = a - a = 0 \]
This leads us to the conclusion that:
\[ c = a + b \]

### Final Answer:
The relationship between \( a \), \( b \), and \( c \) is:
\[ c = a + b \]