In `triangleABC,` if `a=2, b=1, c=sqrt(3)`, then `angleA=`

Options

A. `90^(@)`
B. `60^(@)`
C. `30^(@)`
D. `45^(@)`

Answer

To find angle A in triangle ABC where \( a = 2 \), \( b = 1 \), and \( c = \sqrt{3} \), we can use the cosine rule. The cosine rule states that:

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

### Step 1: Substitute the values into the cosine formula.

Given:
- \( a = 2 \)
- \( b = 1 \)
- \( c = \sqrt{3} \)

We substitute these values into the formula:

\[ \cos A = \frac{1^2 + (\sqrt{3})^2 - 2^2}{2 \cdot 1 \cdot \sqrt{3}} \]

### Step 2: Calculate the squares.

Calculating the squares:

- \( 1^2 = 1 \)
- \( (\sqrt{3})^2 = 3 \)
- \( 2^2 = 4 \)

Now substitute these values back into the equation:

\[ \cos A = \frac{1 + 3 - 4}{2 \cdot 1 \cdot \sqrt{3}} \]

### Step 3: Simplify the numerator.

Now simplify the numerator:

\[ \cos A = \frac{0}{2 \cdot \sqrt{3}} = 0 \]

### Step 4: Determine angle A.

Since \( \cos A = 0 \), we find angle A:

\[ A = \cos^{-1}(0) \]

The angle whose cosine is 0 is \( 90^\circ \) (or \( \frac{\pi}{2} \) radians).

### Final Answer:

Thus, angle A is \( 90^\circ \).

---