Question
In a $$\vartriangle ABC,A = \frac{{2\pi }}{3},b - c = 3\sqrt 3 \,cm$$ and $${\text{ar}}\left( {\vartriangle ABC} \right) = \frac{{9\sqrt 3 }}{2}\,c{m^2}.$$ Then $$a$$ is
A.
$$6\sqrt 3 \,cm$$
B.
$$9\,cm$$
C.
$$18\,cm$$
D.
None of these
Answer :
$$9\,cm$$
Solution :
$$\eqalign{
& \frac{1}{2}bc\sin \frac{{2\pi }}{3} = \frac{{9\sqrt 3 }}{2}\,\,{\text{or, }}\frac{1}{2} \cdot \frac{{\sqrt 3 }}{2} \cdot bc = \frac{{9\sqrt 3 }}{2} \cr
& \Rightarrow \,\,bc = 18. \cr
& {\text{Also, }}\cos \frac{{2\pi }}{3} = \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}} \cr
& \Rightarrow \,\, - \frac{1}{2} = \frac{{{{\left( {b - c} \right)}^2} + 2bc - {a^2}}}{{2bc}} \cr
& {\text{or, }}{\left( {b - c} \right)^2} + 3bc - {a^2} = 0\,\,\,\,{\text{or, }}27 + 54 = {a^2}. \cr} $$