Given `b = 2, c = sqrt3, angle A = 30^(@)`, then inradius of `DeltaABC` is

Options

A. `(sqrt3 -1)/(2)`
B. `(sqrt3 + 1)/(2)`
C. `(sqrt3-1)/(4)`
D. none of these

Explanation

To find the inradius of triangle ABC given \( b = 2 \), \( c = \sqrt{3} \), and \( \angle A = 30^\circ \), we can follow these steps:

### Step 1: Calculate side \( a \)
We will use the cosine rule to find side \( a \):
\[ a = \sqrt{b^2 + c^2 - 2bc \cos A} \]
Substituting the values:
\[ a = \sqrt{2^2 + (\sqrt{3})^2 - 2 \cdot 2 \cdot \sqrt{3} \cdot \cos(30^\circ)} \]
We know that \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \):
\[ a = \sqrt{4 + 3 - 2 \cdot 2 \cdot \sqrt{3} \cdot \frac{\sqrt{3}}{2}} \]
\[ = \sqrt{4 + 3 - 2 \cdot 2 \cdot \frac{3}{2}} \]
\[ = \sqrt{4 + 3 - 6} = \sqrt{1} = 1 \]

### Step 2: Calculate the semi-perimeter \( s \)
The semi-perimeter \( s \) is given by:
\[ s = \frac{a + b + c}{2} \]
Substituting the values:
\[ s = \frac{1 + 2 + \sqrt{3}}{2} = \frac{3 + \sqrt{3}}{2} \]

### Step 3: Calculate the inradius \( r \)
The formula for the inradius \( r \) is:
\[ r = \frac{A}{s} \]
where \( A \) is the area of the triangle. We can use the formula:
\[ A = \frac{1}{2}bc \sin A \]
Substituting the values:
\[ A = \frac{1}{2} \cdot 2 \cdot \sqrt{3} \cdot \sin(30^\circ) \]
We know that \( \sin(30^\circ) = \frac{1}{2} \):
\[ A = \frac{1}{2} \cdot 2 \cdot \sqrt{3} \cdot \frac{1}{2} = \frac{\sqrt{3}}{2} \]

Now substituting \( A \) and \( s \) into the inradius formula:
\[ r = \frac{\frac{\sqrt{3}}{2}}{\frac{3 + \sqrt{3}}{2}} = \frac{\sqrt{3}}{3 + \sqrt{3}} \]

### Step 4: Rationalizing the denominator
To simplify \( r \):
\[ r = \frac{\sqrt{3}(3 - \sqrt{3})}{(3 + \sqrt{3})(3 - \sqrt{3})} = \frac{3\sqrt{3} - 3}{9 - 3} = \frac{3\sqrt{3} - 3}{6} = \frac{\sqrt{3} - 1}{2} \]

Thus, the inradius \( r \) of triangle ABC is:
\[ r = \frac{\sqrt{3} - 1}{2} \]

### Final Answer
The inradius of triangle ABC is \( \frac{\sqrt{3} - 1}{2} \).

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