Given b=2 ,` c= sqrt(3), angle A= 30^@` then inradius of `Delta ABC` is :-

Options

A. `(sqrt(3)-1)/(2)`
B. `(sqrt(3)+1)/(2)`
C. `(sqrt(3)-1)/(4)`
D. Non of these

Answer

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

### Step 1: Calculate the Area (Δ) of Triangle ABC
The area of a triangle can be calculated using the formula:
\[ \Delta = \frac{1}{2} \times b \times c \times \sin A \]
Substituting the values:
\[ \Delta = \frac{1}{2} \times 2 \times \sqrt{3} \times \sin(30^\circ) \]
Since \( \sin(30^\circ) = \frac{1}{2} \):
\[ \Delta = \frac{1}{2} \times 2 \times \sqrt{3} \times \frac{1}{2} = \frac{\sqrt{3}}{2} \]

### Step 2: Use the Cosine Rule to Find Side a
Using the cosine rule:
\[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} \]
Substituting the known values:
\[ \cos(30^\circ) = \frac{2^2 + (\sqrt{3})^2 - a^2}{2 \times 2 \times \sqrt{3}} \]
Calculating \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \):
\[ \frac{\sqrt{3}}{2} = \frac{4 + 3 - a^2}{4\sqrt{3}} \]
Cross-multiplying gives:
\[ 2\sqrt{3} \cdot \sqrt{3} = 7 - a^2 \]
This simplifies to:
\[ 6 = 7 - a^2 \implies a^2 = 1 \implies a = 1 \]

### Step 3: 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 4: Calculate the Inradius (r)
The inradius \( r \) is given by:
\[ r = \frac{\Delta}{s} \]
Substituting the values of \( \Delta \) and \( s \):
\[ r = \frac{\frac{\sqrt{3}}{2}}{\frac{3 + \sqrt{3}}{2}} = \frac{\sqrt{3}}{3 + \sqrt{3}} \]
To rationalize the denominator, multiply the numerator and denominator by \( 3 - \sqrt{3} \):
\[ r = \frac{\sqrt{3}(3 - \sqrt{3})}{(3 + \sqrt{3})(3 - \sqrt{3})} = \frac{3\sqrt{3} - 3}{9 - 3} = \frac{3\sqrt{3} - 3}{6} \]
This simplifies to:
\[ r = \frac{\sqrt{3} - 1}{2} \]

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