If `cos 3A +cos 3B+cos 3C=1` then one of the angles of the triangle ABC is

Options

A. `60^@`
B. `90^@`
C. `120^@`
D. `30^@`

Answer

To solve the problem where \( \cos 3A + \cos 3B + \cos 3C = 1 \) and find one of the angles of triangle ABC, we can follow these steps:

### Step 1: Understand the triangle angle sum
Since \( A, B, C \) are angles of a triangle, we know that:
\[ A + B + C = \pi \quad \text{(or 180 degrees)} \]

### Step 2: Rewrite one angle in terms of the others
We can express \( C \) in terms of \( A \) and \( B \):
\[ C = \pi - A - B \]

### Step 3: Substitute \( C \) into the equation
Now substitute \( C \) into the equation \( \cos 3A + \cos 3B + \cos 3C = 1 \):
\[ \cos 3A + \cos 3B + \cos(3(\pi - A - B)) = 1 \]
Using the cosine identity \( \cos(\pi - x) = -\cos x \):
\[ \cos 3A + \cos 3B - \cos(3A + 3B) = 1 \]

### Step 4: Use the cosine addition formula
We can use the cosine addition formula:
\[ \cos(3A + 3B) = \cos 3A \cos 3B - \sin 3A \sin 3B \]
Thus, we rewrite our equation:
\[ \cos 3A + \cos 3B - (\cos 3A \cos 3B - \sin 3A \sin 3B) = 1 \]

### Step 5: Simplify the equation
Rearranging gives:
\[ \cos 3A + \cos 3B - \cos 3A \cos 3B + \sin 3A \sin 3B = 1 \]
This can be complex to simplify directly, so we will analyze the implications of the original equation.

### Step 6: Analyze the equation
The equation \( \cos 3A + \cos 3B + \cos 3C = 1 \) suggests that the angles \( 3A, 3B, 3C \) must be such that their cosines sum to 1.

### Step 7: Consider special angles
One possible case is when one of the angles is \( 60^\circ \) (or \( \frac{\pi}{3} \)). Let's assume \( A = 60^\circ \):
\[ 3A = 180^\circ \implies \cos 3A = \cos 180^\circ = -1 \]
Then, we need to check if \( \cos 3B + \cos 3C = 2 \) can hold true. The only way for this to happen is if both \( B \) and \( C \) are \( 60^\circ \) as well.

### Conclusion
Thus, if \( A = 60^\circ \), then \( B + C = 120^\circ \) and both can also be \( 60^\circ \). Therefore, one of the angles of triangle ABC is:
\[ \boxed{60^\circ} \]