Question
Two sides of a triangle are given by the roots of the equation $${x^2} - 2\sqrt 3 x + 2 = 0.$$ The angle between the sides is $$\frac{\pi }{3}.$$ The perimeter of the triangle is
A.
$$6 + \sqrt 3 $$
B.
$$2\sqrt 3 + \sqrt 6 $$
C.
$$2\sqrt 3 + \sqrt 10 $$
D.
None of these
Answer :
$$2\sqrt 3 + \sqrt 6 $$
Solution :
Here, $$a + b = 2\sqrt 3 ,ab = 2\,{\text{and }}C = \frac{\pi }{3}.$$
$$\eqalign{
& \cos C = \frac{{{a^2} + {b^2} - {c^2}}}{{2ab}} \cr
& \Rightarrow \,\,\frac{1}{2} = \frac{{{a^2} + {b^2} - {c^2}}}{{2ab}} \cr
& \Rightarrow \,\,{a^2} + {b^2} - {c^2} = ab\,\,{\text{or, }}{\left( {a + b} \right)^2} - 2ab - {c^2} = ab \cr
& {\text{or, }}12 - 4 - {c^2} = 2\,\,\,{\text{or, }}c = \sqrt 6 . \cr} $$
∴ the perimeter $$ = a + b + c = 2\sqrt 3 + \sqrt 6 .$$