Question

The perimeter of a $\triangle$ ABC is 6 times the arithmetic mean of the sines of its angles. If the side a is 1, then the angle A is

• A. π/6
• B. π/3
• C. π/2
• D. π

Answer 1

Answer: (A)

π/6

Answer 2

The perimeter of $\triangle ABC$ is $a+b+c$. The arithmetic mean of the sines of its angles is $\frac{\sin A + \sin B + \sin C}{3}$.

Given $a+b+c = 6 \times \frac{\sin A + \sin B + \sin C}{3}$, which simplifies to $a+b+c = 2(\sin A + \sin B + \sin C)$.

By the Sine Rule, $a = 2R \sin A$, $b = 2R \sin B$, $c = 2R \sin C$, where R is the circumradius.

Substituting these into the equation: $2R \sin A + 2R \sin B + 2R \sin C = 2(\sin A + \sin B + \sin C)$ $2R (\sin A + \sin B + \sin C) = 2(\sin A + \sin B + \sin C)$

Since $\sin A + \sin B + \sin C \neq 0$ for a triangle, we can divide both sides by this sum, yielding $2R = 2$, so $R=1$.

Given side $a=1$. Using the Sine Rule again: $\frac{a}{\sin A} = 2R$.

Substituting $a=1$ and $R=1$: $\frac{1}{\sin A} = 2(1)$, which gives $\sin A = \frac{1}{2}$.

For an angle in a triangle, $A = \frac{\pi}{6}$ or $A = \frac{5\pi}{6}$. Given the options, $\frac{\pi}{6}$ is the correct answer.