Here   AB = x

and   BC = y

and   $$\lambda $$ = linear mass density.

As centre of mass is below point A, so horizontal distance of the centre of mass from B is = xcos60^o = $${x \over 2}$$

$$ \therefore $$   X_CM = $${x \over 2}$$ = $${{{m_1}{x_1} + {m_2}{x_2}} \over {{m_1} + {m_2}}}$$

$$ \Rightarrow $$   $${x \over 2}$$ = $${{\left( {\lambda x} \right)\left( {{x \over 2}} \right)\cos {{60}^o} + \left( {\lambda y} \right)\left( {{y \over 2}} \right)} \over {\lambda \left( {x + y} \right)}}$$

$$ \Rightarrow $$   $${x \over 2}$$ = $${{{{{x^2}} \over 4} + {{{y^2}} \over 2}} \over {x + y}}$$

$$ \Rightarrow $$   x² + xy = $${{{x^2}} \over 2} + {y^2}$$

$$ \Rightarrow $$   x² + 2xy $$-$$ 2y² = 0

$$ \therefore $$   x = $${{ - 2y \pm \sqrt {{{\left( {2y} \right)}^2} - 4.1\left( { - 2{y^2}} \right)} } \over {2.1}}$$

=    $${{ - 2y \pm \sqrt {12{y^2}} } \over 2}$$

=   $$-$$ y $$ \pm $$ $$\sqrt 3 $$y

$${x \over y} \ne - \sqrt 3 - 1$$  as  $${x \over y}$$ = positive.

$$ \therefore $$   $${x \over y}$$ = $$\sqrt 3 - 1$$

$$ \Rightarrow $$   $${y \over x}$$ = $${1 \over {\sqrt 3 - 1}} \times {{\sqrt 3 + 1} \over {\sqrt 3 + 1}}$$

=   $${{\sqrt 3 + 1} \over 2}$$

=   $${{2.732} \over 2}$$

=   1.366 $$ \simeq $$ 1.37