In a `DeltaABC`, if a = 3, `b=2sqrt(3)` and `c=sqrt(3)`, then A=

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. `75^(@)`

Explanation

To find the angle \( A \) in triangle \( ABC \) where \( a = 3 \), \( b = 2\sqrt{3} \), and \( c = \sqrt{3} \), we can use the Law of Cosines. The Law of Cosines states that:

\[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} \]

### Step 1: Substitute the values of \( a \), \( b \), and \( c \)

Given:
- \( a = 3 \)
- \( b = 2\sqrt{3} \)
- \( c = \sqrt{3} \)

Substituting these values into the formula:

\[ \cos A = \frac{(2\sqrt{3})^2 + (\sqrt{3})^2 - (3)^2}{2 \cdot (2\sqrt{3}) \cdot (\sqrt{3})} \]

### Step 2: Calculate \( b^2 \), \( c^2 \), and \( a^2 \)

Calculating each term:

\[ b^2 = (2\sqrt{3})^2 = 4 \cdot 3 = 12 \]
\[ c^2 = (\sqrt{3})^2 = 3 \]
\[ a^2 = (3)^2 = 9 \]

### Step 3: Substitute the squared values into the equation

Now substituting these squared values back into the equation:

\[ \cos A = \frac{12 + 3 - 9}{2 \cdot (2\sqrt{3}) \cdot (\sqrt{3})} \]

### Step 4: Simplify the numerator

Calculating the numerator:

\[ 12 + 3 - 9 = 6 \]

### Step 5: Calculate the denominator

Now calculate the denominator:

\[ 2 \cdot (2\sqrt{3}) \cdot (\sqrt{3}) = 2 \cdot 2 \cdot 3 = 12 \]

### Step 6: Substitute back into the equation

Now we have:

\[ \cos A = \frac{6}{12} = \frac{1}{2} \]

### Step 7: Find angle \( A \)

To find angle \( A \):

\[ A = \cos^{-1}\left(\frac{1}{2}\right) \]

From trigonometric values, we know:

\[ A = 60^\circ \]

### Final Answer

Thus, the angle \( A \) is \( 60^\circ \).

---