In the triangle ABC.`b=sqrt3cm,c=1cm.angleA=30^(@)`. What is the value of a?

Options

A. `sqrt2`
B. 2cm
C. 1cm
D. `(1)/(2)cm`

Answer

To find the value of side \( a \) in triangle \( ABC \) where \( b = \sqrt{3} \) cm, \( c = 1 \) cm, and \( \angle A = 30^\circ \), we can use the cosine rule. The cosine rule states:

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

### Step-by-Step Solution:

1. **Identify the known values:**
- \( b = \sqrt{3} \) cm
- \( c = 1 \) cm
- \( \angle A = 30^\circ \)

2. **Substitute the values into the cosine rule:**
\[ \cos(30^\circ) = \frac{(\sqrt{3})^2 + (1)^2 - a^2}{2 \cdot \sqrt{3} \cdot 1} \]

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

4. **Substitute \( \cos(30^\circ) \) into the equation:**
\[ \frac{\sqrt{3}}{2} = \frac{3 + 1 - a^2}{2\sqrt{3}} \]

5. **Cross-multiply to eliminate the fraction:**
\[ \sqrt{3} \cdot \sqrt{3} = 2(4 - a^2) \]
\[ 3 = 8 - 2a^2 \]

6. **Rearrange the equation to solve for \( a^2 \):**
\[ 2a^2 = 8 - 3 \]
\[ 2a^2 = 5 \]
\[ a^2 = \frac{5}{2} \]

7. **Take the square root to find \( a \):**
\[ a = \sqrt{\frac{5}{2}} = \frac{\sqrt{10}}{2} \]

8. **Since \( a \) must be positive, we discard the negative root:**
\[ a = \frac{\sqrt{10}}{2} \text{ cm} \]

### Final Answer:
The value of \( a \) is \( \frac{\sqrt{10}}{2} \) cm.