Question
Two sides of a triangle are $$2\sqrt 2 \,cm$$ and $$2\sqrt 3 \,cm$$ and the angle opposite to the shorter side of the two is $$\frac{\pi }{4}.$$ The largest possible length of the third side is
A.
$$\sqrt 2 \left( {\sqrt 3 + 1} \right)\,cm$$
B.
$$\left( {6 + \sqrt 2 } \right)\,cm$$
C.
$$\left( {\sqrt 6 - \sqrt 2 } \right)\,cm$$
D.
None of these
Answer :
$$\sqrt 2 \left( {\sqrt 3 + 1} \right)\,cm$$
Solution :
$$\eqalign{
& \cos \frac{\pi }{4} = \frac{{{{\left( {2\sqrt 3 } \right)}^2} + {x^2} - {{\left( {2\sqrt 2 } \right)}^2}}}{{2 \cdot 2\sqrt 3 \cdot x}}\,\,{\text{or, }}\frac{{4\sqrt 3 }}{{\sqrt 2 }}x = {x^2} + 4\,\,{\text{or, }}{x^2} - \frac{{4\sqrt 3 }}{{\sqrt 2 }}x + 4 = 0. \cr
& \therefore \,\,x = \sqrt 6 \pm \sqrt 2 . \cr} $$