If in a triangle `ABC,a^2+b^2+c^2 = ca+ ab sqrt3`, then the triangle is

Options

A. equilateral
B. right-angled and isosceles
C. right angled with `A = 90^(@), B = 60^(@), C = 30^(@)`
D. None of these

Answer

To determine the type of triangle given the equation \( a^2 + b^2 + c^2 = ca + ab \sqrt{3} \), we will follow these steps:

### Step 1: Rewrite the Given Equation
We start with the equation:
\[ a^2 + b^2 + c^2 = ca + ab \sqrt{3} \]

### Step 2: Rearranging the Equation
Rearranging gives:
\[ a^2 + b^2 + c^2 - ca - ab \sqrt{3} = 0 \]

### Step 3: Use the Law of Sines
Using the Law of Sines, we can express the sides in terms of angles:
\[ a = k \sin A, \quad b = k \sin B, \quad c = k \sin C \]
where \( k \) is a constant.

### Step 4: Substitute into the Equation
Substituting these into our rearranged equation:
\[ (k \sin A)^2 + (k \sin B)^2 + (k \sin C)^2 - (k \sin C)(k \sin A) - (k \sin A)(k \sin B) \sqrt{3} = 0 \]
This simplifies to:
\[ k^2 (\sin^2 A + \sin^2 B + \sin^2 C) - k^2 (\sin C \sin A + \sqrt{3} \sin A \sin B) = 0 \]

### Step 5: Factor Out \( k^2 \)
Factoring out \( k^2 \) gives:
\[ k^2 \left( \sin^2 A + \sin^2 B + \sin^2 C - (\sin C \sin A + \sqrt{3} \sin A \sin B) \right) = 0 \]
Since \( k^2 \neq 0 \), we can set the inside of the parentheses to zero:
\[ \sin^2 A + \sin^2 B + \sin^2 C = \sin C \sin A + \sqrt{3} \sin A \sin B \]

### Step 6: Recognize the Form
Notice that the right-hand side can be related to the cosine of angles:
\[ \cos B = \frac{1}{2}, \quad \cos C = \frac{\sqrt{3}}{2} \]
This implies:
\[ B = \frac{\pi}{3}, \quad C = \frac{\pi}{6} \]

### Step 7: Find Angle A
Using the fact that the sum of angles in a triangle is \( \pi \):
\[ A + B + C = \pi \]
Substituting the values of \( B \) and \( C \):
\[ A + \frac{\pi}{3} + \frac{\pi}{6} = \pi \]
This simplifies to:
\[ A + \frac{\pi}{2} = \pi \implies A = \frac{\pi}{2} \]

### Conclusion
Since \( A = \frac{\pi}{2} \), we conclude that triangle \( ABC \) is a right-angled triangle.