P is a point inside the equilateral traingle ABC such that `PA=1cm, PB=sqrt(2)cm` and `PC=sqrt(3)`. Then the side of the traingle `("in cm")` equal.

Options

A. `sqrt(6-2sqrt(3))`
B. `sqrt(6-sqrt(3))`
C. `sqrt(3+sqrt(6))`
D. `sqrt(3+2sqrt(6))`

Explanation

To find the side length of the equilateral triangle \( ABC \) given the distances from point \( P \) to the vertices \( A \), \( B \), and \( C \), we can use the following steps:

### Step 1: Set up the problem
Let \( PA = 1 \, \text{cm} \), \( PB = \sqrt{2} \, \text{cm} \), and \( PC = \sqrt{3} \, \text{cm} \). Let the side length of the equilateral triangle \( ABC \) be \( a \).

### Step 2: Apply the Cosine Rule
Using the Cosine Rule in triangles \( PAB \) and \( PAC \):

1. In triangle \( PAB \):
\[ AB^2 = PA^2 + PB^2 - 2 \cdot PA \cdot PB \cdot \cos(\theta) \]
where \( \theta \) is the angle \( \angle APB \).

Therefore:
\[ a^2 = 1^2 + (\sqrt{2})^2 - 2 \cdot 1 \cdot \sqrt{2} \cdot \cos(\theta) \]
Simplifying this gives:
\[ a^2 = 1 + 2 - 2\sqrt{2} \cos(\theta) \]
\[ a^2 = 3 - 2\sqrt{2} \cos(\theta) \quad \text{(1)} \]

2. In triangle \( PAC \):
\[ AC^2 = PA^2 + PC^2 - 2 \cdot PA \cdot PC \cdot \cos(60^\circ - \theta) \]
Since \( \cos(60^\circ) = \frac{1}{2} \):
\[ a^2 = 1^2 + (\sqrt{3})^2 - 2 \cdot 1 \cdot \sqrt{3} \cdot \left(\frac{1}{2} \cos(\theta) + \frac{\sqrt{3}}{2} \sin(\theta)\right) \]
Simplifying this gives:
\[ a^2 = 1 + 3 - \sqrt{3} \cos(\theta) - \sqrt{3} \sin(\theta) \quad \text{(2)} \]

### Step 3: Set equations equal
From equations (1) and (2), we have:
\[ 3 - 2\sqrt{2} \cos(\theta) = 4 - \sqrt{3} \cos(\theta) - \sqrt{3} \sin(\theta) \]

### Step 4: Rearranging the equation
Rearranging gives:
\[ \sqrt{3} \cos(\theta) + \sqrt{3} \sin(\theta) - 2\sqrt{2} \cos(\theta) = 1 \]
Factoring out \( \cos(\theta) \):
\[ (\sqrt{3} - 2\sqrt{2}) \cos(\theta) + \sqrt{3} \sin(\theta) = 1 \]

### Step 5: Solve for \( a \)
To find \( a \), we can use the derived equations. We can also use the fact that in an equilateral triangle, the sum of the squares of the distances from any interior point to the vertices is equal to the sum of the squares of the sides.

Using the relation:
\[ PA^2 + PB^2 + PC^2 = a^2 + a^2 + a^2 \]
Substituting the known values:
\[ 1^2 + (\sqrt{2})^2 + (\sqrt{3})^2 = 3a^2 \]
\[ 1 + 2 + 3 = 3a^2 \]
\[ 6 = 3a^2 \]
\[ a^2 = 2 \quad \Rightarrow \quad a = \sqrt{2} \]

### Final Answer
The side length of the triangle \( ABC \) is \( \sqrt{2} \, \text{cm} \).