In a triangle `ABC, a=1cm, b=sqrt3 and C=(pi)/(6)`, find the third side of the triangle :

Options

A. 1 cm
B. 1.5 cm
C. 2 cm
D. `3sqrt3cm`

Answer

To find the third side \( c \) of triangle \( ABC \) where \( a = 1 \, \text{cm} \), \( b = \sqrt{3} \, \text{cm} \), and \( C = \frac{\pi}{6} \) radians (or \( 30^\circ \)), we will use the cosine rule.

### Step-by-step Solution:

1. **Identify the known values**:
- \( a = 1 \, \text{cm} \)
- \( b = \sqrt{3} \, \text{cm} \)
- \( C = 30^\circ \)

2. **Write the cosine rule formula**:
The cosine rule states:
\[ c^2 = a^2 + b^2 - 2ab \cos(C) \]

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

4. **Substitute the known values into the cosine rule**:
\[ c^2 = 1^2 + (\sqrt{3})^2 - 2 \cdot 1 \cdot \sqrt{3} \cdot \frac{\sqrt{3}}{2} \]

5. **Calculate each term**:
- \( 1^2 = 1 \)
- \( (\sqrt{3})^2 = 3 \)
- \( 2 \cdot 1 \cdot \sqrt{3} \cdot \frac{\sqrt{3}}{2} = \sqrt{3} \cdot \sqrt{3} = 3 \)

6. **Combine the terms**:
\[ c^2 = 1 + 3 - 3 \]
\[ c^2 = 1 \]

7. **Take the square root to find \( c \)**:
\[ c = \sqrt{1} = 1 \, \text{cm} \]

### Final Answer:
The third side \( c \) of triangle \( ABC \) is \( 1 \, \text{cm} \).

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