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: Read the condition.
The perimeter of triangle $ABC$ equals $6$ times the arithmetic mean of the sines of its angles, and $a=1$. We must find angle $A$.
Step 2: Write the perimeter and the mean.
The perimeter is $a+b+c$. The arithmetic mean of the three sines is $\frac{\sin A + \sin B + \sin C}{3}$. The condition says \[ a+b+c = 6\cdot\frac{\sin A + \sin B + \sin C}{3} = 2(\sin A + \sin B + \sin C). \]
Step 3: Use the sine rule.
The sine rule says $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$. So $a = 2R\sin A$, and similarly for $b$ and $c$. Then \[ a+b+c = 2R(\sin A + \sin B + \sin C). \]
Step 4: Compare the two expressions.
We now have $2R(\sin A+\sin B+\sin C) = 2(\sin A+\sin B+\sin C)$. Cancelling the common sum gives $2R = 2$, so $R = 1$.
Step 5: Use $a = 1$ in the sine rule.
From $\frac{a}{\sin A} = 2R$ we get $\sin A = \frac{a}{2R} = \frac{1}{2}$.
Step 6: Find the angle.
The angle in a triangle with $\sin A = \frac{1}{2}$ that fits is $A = \frac{\pi}{6}$. \[ \boxed{A = \dfrac{\pi}{6}} \]