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_____.

Answer 1

Find the angle.

Let the sides of the triangle are $a,b\mathrm{and}c$.

It is given that the perimeter of a triangle $ABC$ is $6$ times the Arithmetic Mean of the sines of its angles.

$$\begin{gathered} \therefore a+b+c=6\left(\frac{\sin A+\sin B+\sin C}{3}\right) \\ a+b+c=2(\sin A+\sin B+\sin C)...\left(1\right) \end{gathered}$$

From the law of sine,

$$\begin{gathered} \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k \\ \Rightarrow a=k\sin A \\ \Rightarrow b=k\sin B \\ \Rightarrow c=k\sin C \\ \therefore a+b+c=k(\sin A+\sin B+\sin C)...\left(2\right) \end{gathered}$$

Compare equation $1$ and $2$.

We get, $k=2$

$$\begin{gathered} \therefore \frac{a}{\sin A}=2 \\ \Rightarrow \frac{1}{\sin A}=2\mathbf{}\mathbf{(}\mathbf{\because }\mathit{a}\mathbf{=}\mathbf{1}\mathbf{)} \\ \Rightarrow \sin A=\frac{1}{2} \\ \Rightarrow A={\sin }^{-1}\left(\frac{1}{2}\right) \\ \Rightarrow A=\frac{\mathrm{\pi }}{6} \end{gathered}$$

Hence, the angle $A$is $\frac{\mathrm{\pi }}{6}$.