In `DeltaABC, tan B + tanC=5` and `tan AtanC=3`, then

Options

A. `DeltaABC` is an acute angled triangle
B. `DeltaABC` is an obtuse angled triangle
C. sum of all possible values of `tanA` is 10
D. sum of all possible values of tan A is 9

Answer

To solve the problem, we will use the given information about triangle \( ABC \) and the relationships between the tangents of the angles.

### Step-by-Step Solution:

1. **Write down the given equations:**
We are given:
\[ \tan B + \tan C = 5 \quad \text{(1)} \]
\[ \tan A \tan C = 3 \quad \text{(2)} \]

2. **Use the identity for the sum of tangents in a triangle:**
In any triangle, the following identity holds:
\[ \tan A + \tan B + \tan C = \tan A \tan B \tan C \quad \text{(3)} \]

3. **Substitute the known values into the identity:**
From equation (1), we can express \( \tan B + \tan C \) as 5. Thus, we can rewrite equation (3) as:
\[ \tan A + 5 = \tan A \tan B \tan C \]

4. **Express \( \tan B \) in terms of \( \tan C \):**
From equation (1), we can express \( \tan B \):
\[ \tan B = 5 - \tan C \quad \text{(4)} \]

5. **Substitute \( \tan B \) into the identity:**
Substitute equation (4) into equation (3):
\[ \tan A + 5 = \tan A (5 - \tan C) \tan C \]

6. **Substitute \( \tan C \) from equation (2):**
From equation (2), we have \( \tan C = \frac{3}{\tan A} \). Substitute this into the equation:
\[ \tan A + 5 = \tan A \left(5 - \frac{3}{\tan A}\right) \cdot \frac{3}{\tan A} \]

7. **Simplify the equation:**
This simplifies to:
\[ \tan A + 5 = \frac{3(5\tan A - 3)}{\tan A} \]
Multiply through by \( \tan A \) to eliminate the fraction:
\[ \tan^2 A + 5\tan A = 15\tan A - 9 \]

8. **Rearrange to form a quadratic equation:**
Rearranging gives us:
\[ \tan^2 A - 10\tan A + 9 = 0 \]

9. **Solve the quadratic equation:**
We can use the quadratic formula \( \tan A = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = -10, c = 9 \):
\[ \tan A = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 1 \cdot 9}}{2 \cdot 1} \]
\[ \tan A = \frac{10 \pm \sqrt{100 - 36}}{2} = \frac{10 \pm \sqrt{64}}{2} = \frac{10 \pm 8}{2} \]
This gives us two solutions:
\[ \tan A = \frac{18}{2} = 9 \quad \text{or} \quad \tan A = \frac{2}{2} = 1 \]

10. **Determine \( \tan B \) and \( \tan C \) for each case:**
- If \( \tan A = 9 \):
\[ \tan C = \frac{3}{9} = \frac{1}{3} \]
\[ \tan B = 5 - \tan C = 5 - \frac{1}{3} = \frac{15 - 1}{3} = \frac{14}{3} \]

- If \( \tan A = 1 \):
\[ \tan C = \frac{3}{1} = 3 \]
\[ \tan B = 5 - \tan C = 5 - 3 = 2 \]

11. **Check the angles:**
- For \( \tan A = 9 \), \( \tan B = \frac{14}{3} \), \( \tan C = \frac{1}{3} \): All angles are acute.
- For \( \tan A = 1 \), \( \tan B = 2 \), \( \tan C = 3 \): All angles are acute.

### Final Answers:
The possible values of \( \tan A \) are \( 1 \) and \( 9 \). The sum of all possible values of \( \tan A \) is \( 10 \).