Question

With usual notations, perimeter of a triangle \( ABC \) is 6 times the arithmetic mean of sine of its angles. If \( a = 1 \), then measure of angle \( A = \)

• A. \( \frac{\pi}{3} \)
• B. \( \frac{\pi}{2} \)
• C. \( \frac{\pi}{4} \)
• D. \( \frac{\pi}{6} \)

Hint:

In any triangle, \(a = 2R \sin A\). Perimeter = \(2R(\sin A + \sin B + \sin C)\). Use given condition to find \(R\), then find \(A\).

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
Perimeter \(P = a + b + c\). Arithmetic mean of sines = \(\frac{\sin A + \sin B + \sin C}{3}\). Given \(P = 6 \times \frac{\sin A + \sin B + \sin C}{3} = 2(\sin A + \sin B + \sin C)\). Also \(a = 1\).

Step 2: Key Formula or Approach:
Use sine rule: \(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R\). Then \(a + b + c = 2R(\sin A + \sin B + \sin C)\).

Step 3: Detailed Explanation:
From sine rule, \(b = 2R \sin B\), \(c = 2R \sin C\), and \(a = 2R \sin A = 1\).
Perimeter: \(a + b + c = 2R(\sin A + \sin B + \sin C)\).
Given \(a + b + c = 2(\sin A + \sin B + \sin C)\).
Equating: \(2R(\sin A + \sin B + \sin C) = 2(\sin A + \sin B + \sin C)\).
Assuming \(\sin A + \sin B + \sin C \neq 0\), we get \(2R = 2 \implies R = 1\).
Then \(a = 2R \sin A = 2 \sin A = 1 \implies \sin A = \frac{1}{2}\).
Thus \(A = \frac{\pi}{6}\) (since \(A\) is an angle of a triangle, \(A = \frac{5\pi}{6}\) is not possible).

Step 4: Final Answer:
Option (D) is correct.