Question

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

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

Hint:

In problems involving the perimeter and angles of a triangle, using trigonometric identities and symmetry can simplify calculations significantly.

Answer 1

The Correct Option is B

Step 1: Understanding the condition.
We are given that the perimeter of triangle ABC is 6 times the arithmetic mean of the sines of its angles. The perimeter \( P \) of triangle ABC is: \[ P = a + b + c. \]
The arithmetic mean of the sines of the angles is: \[ \frac{\sin A + \sin B + \sin C}{3}. \]

Step 2: Equation setup.
The problem tells us that:
\[ P = 6 \times \frac{\sin A + \sin B + \sin C}{3}. \]
Substitute \( P = a + b + c \) into the equation:
\[ a + b + c = 2 (\sin A + \sin B + \sin C). \]

Step 3: Substituting \( a = 1 \) and simplifying.
We know that \( a = 1 \). So the equation becomes:
\[ 1 + b + c = 2 (\sin A + \sin B + \sin C). \]

Step 4: Using known identities.
We know that \( \sin A + \sin B + \sin C \) is maximized when \( A = B = C = \frac{\pi}{3} \). Substituting \( A = \frac{\pi}{3} \) into the equation, we find that:
\[ \sin A + \sin B + \sin C = 3 \times \sin \frac{\pi}{3} = 3 \times \frac{\sqrt{3}}{2}. \]

Step 5: Solving for \( \angle A \).
Solving the equation for \( A \), we get \( \angle A = \frac{\pi}{3} \).
Final Answer:
The correct value of \( \angle A \) is: \[ \boxed{\frac{\pi}{3}}. \]