Question
In a $$\vartriangle ABC,a = 1$$ and the perimeter is six times the AM of the sines of the angles. The measure of $$\angle A$$ is
A.
$$\frac{\pi }{3}$$
B.
$$\frac{\pi }{2}$$
C.
$$\frac{\pi }{6}$$
D.
$$\frac{\pi }{4}$$
Answer :
$$\frac{\pi }{6}$$
Solution :
$$\eqalign{
& 1 + b + c = 6.\frac{{\sin A + \sin B + \sin C}}{3} = 2\left( {\frac{1}{{2R}} + \frac{b}{{2R}} + \frac{c}{{2R}}} \right) = \frac{1}{R}\left( {1 + b + c} \right) \cr
& \therefore \,\,R = 1\,\,\,\,\,\left( {\because \,1 + b + c \ne 0} \right). \cr
& {\text{So}},\frac{1}{{\sin A}} = 2R = 2 \cr
& \Rightarrow \,\,A = \frac{\pi }{6}. \cr} $$