In a `DeltaABC`, if a = 40, `c=40sqrt(3)` and `B=30^(@)`, then the triangle is

Options

A. isosceles
B. equilateral
C. right angled
D. none of these

Explanation

To solve the problem, we will use the Law of Cosines to determine the properties of triangle ABC given the sides and angle.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Side \( a = 40 \)
- Side \( c = 40\sqrt{3} \)
- Angle \( B = 30^\circ \)

2. **Apply the Law of Cosines:**
The Law of Cosines states:
\[ \cos B = \frac{a^2 + c^2 - b^2}{2ac} \]
We need to find \( b \) using the values we have.

3. **Substitute the Known Values:**
Substitute \( a \), \( c \), and \( B \) into the equation:
\[ \cos(30^\circ) = \frac{40^2 + (40\sqrt{3})^2 - b^2}{2 \cdot 40 \cdot 40\sqrt{3}} \]

4. **Calculate \( \cos(30^\circ) \):**
We know that:
\[ \cos(30^\circ) = \frac{\sqrt{3}}{2} \]

5. **Calculate \( a^2 \) and \( c^2 \):**
\[ a^2 = 40^2 = 1600 \]
\[ c^2 = (40\sqrt{3})^2 = 1600 \cdot 3 = 4800 \]

6. **Substitute Values into the Equation:**
Now, substituting \( a^2 \) and \( c^2 \) into the equation gives:
\[ \frac{\sqrt{3}}{2} = \frac{1600 + 4800 - b^2}{2 \cdot 40 \cdot 40\sqrt{3}} \]

7. **Simplify the Equation:**
The denominator becomes:
\[ 2 \cdot 40 \cdot 40\sqrt{3} = 3200\sqrt{3} \]
Thus, the equation simplifies to:
\[ \frac{\sqrt{3}}{2} = \frac{6400 - b^2}{3200\sqrt{3}} \]

8. **Cross Multiply:**
Cross multiplying gives:
\[ \sqrt{3} \cdot 3200\sqrt{3} = 2(6400 - b^2) \]
Simplifying this, we have:
\[ 9600 = 12800 - 2b^2 \]

9. **Rearranging to Solve for \( b^2 \):**
Rearranging gives:
\[ 2b^2 = 12800 - 9600 \]
\[ 2b^2 = 3200 \]
\[ b^2 = 1600 \]

10. **Finding \( b \):**
Taking the square root gives:
\[ b = \sqrt{1600} = 40 \]

11. **Determine the Type of Triangle:**
Since \( a = 40 \), \( b = 40 \), and \( c = 40\sqrt{3} \), we see that two sides are equal (\( a = b \)). Therefore, triangle ABC is an **isosceles triangle**.

### Conclusion:
The triangle ABC is an isosceles triangle.