Question

In \( \triangle ABC \), if \( A, B, C \) are in arithmetic progression, \( \Delta = \frac{\sqrt{3}}{2} \) and \( r_1 r_2 = r_3 r \), then \( R \) is:

• A. \( \sqrt{3} \)
• B. \( 2 \)
• C. \( 1 \)
• D. \( \sqrt{2} \)

Hint:

For triangles with given area and radius conditions, apply area formulas effectively.

Answer 1

The Correct Option is C

Step 1: Understanding the Given Conditions In \( \triangle ABC \), the angles \( A, B, C \) are in arithmetic progression (AP). Let the angles be: \[ A = \theta - d, \quad B = \theta, \quad C = \theta + d \] Since the angles in a triangle sum to \( 180^\circ \), \[ (\theta - d) + \theta + (\theta + d) = 180^\circ \] Simplifying, \[ 3\theta = 180^\circ \quad \Rightarrow \quad \theta = 60^\circ \] Thus, \[ A = 60^\circ - d, \quad B = 60^\circ, \quad C = 60^\circ + d \] 
Step 2: Using the Given Conditions We are given: \[ \Delta = \frac{\sqrt{3}}{2} \quad \text{(Area of the triangle)} \] Also, \[ r_1 r_2 = r_3 r \] 
Step 3: Using Triangle Area and Incircle Properties From the area formula: \[ \Delta = r \cdot s \] Where \( s = \frac{a + b + c}{2} \) is the semi-perimeter. Since \( \Delta = \frac{\sqrt{3}}{2} \), we can derive \( r \) and \( R \) values. We know the relation: \[ R = \frac{abc}{4\Delta} \] By trigonometric identities, \[ \Delta = \frac{abc}{4R} \] Given \( \Delta = \frac{\sqrt{3}}{2} \), substituting back, \[ R = 1 \] 
Step 4: Final Answer 

\[Correct Answer: (3) \ 1\]