In a `triangleABC` if 3`angleA=4angleB=6angleC`,calculate `angleA`

Options

A. `angleA=60^(@)`
B. `angleA=80^(@)`
C. `angleA=40^(@)`
D. None of these

Answer

To solve the problem, we need to find the angles \( \angle A \), \( \angle B \), and \( \angle C \) in triangle \( ABC \) given the relationships between them.

### Step-by-Step Solution:

1. **Set up the equations based on the given relationships:**
We have the following relationships:
- \( 3\angle A = 4\angle B \)
- \( 4\angle B = 6\angle C \)

2. **Express \( \angle A \) in terms of \( \angle B \):**
From the first equation, we can express \( \angle A \):
\[ \angle A = \frac{4}{3} \angle B \]

3. **Express \( \angle C \) in terms of \( \angle B \):**
From the second equation, we can express \( \angle C \):
\[ 4\angle B = 6\angle C \implies \angle C = \frac{4}{6} \angle B = \frac{2}{3} \angle B \]

4. **Substitute \( \angle A \) and \( \angle C \) into the triangle angle sum property:**
The sum of angles in a triangle is \( 180^\circ \):
\[ \angle A + \angle B + \angle C = 180^\circ \]
Substituting the expressions for \( \angle A \) and \( \angle C \):
\[ \frac{4}{3} \angle B + \angle B + \frac{2}{3} \angle B = 180^\circ \]

5. **Combine the terms:**
To combine the terms, we can express \( \angle B \) as \( \frac{3}{3} \angle B \):
\[ \frac{4}{3} \angle B + \frac{3}{3} \angle B + \frac{2}{3} \angle B = 180^\circ \]
This simplifies to:
\[ \frac{4 + 3 + 2}{3} \angle B = 180^\circ \]
\[ \frac{9}{3} \angle B = 180^\circ \]
\[ 3\angle B = 180^\circ \]

6. **Solve for \( \angle B \):**
Dividing both sides by 3:
\[ \angle B = 60^\circ \]

7. **Find \( \angle A \) and \( \angle C \):**
Now that we have \( \angle B \), we can find \( \angle A \) and \( \angle C \):
- For \( \angle A \):
\[ \angle A = \frac{4}{3} \times 60^\circ = 80^\circ \]
- For \( \angle C \):
\[ \angle C = \frac{2}{3} \times 60^\circ = 40^\circ \]

8. **Final answer:**
Therefore, the angles are:
- \( \angle A = 80^\circ \)
- \( \angle B = 60^\circ \)
- \( \angle C = 40^\circ \)