We know,

Maximum height of a projectile $$(H) = {{{u^2}{{\sin }^2}\theta } \over {2g}}$$

Given, Ratio of initial velocity of two projectile

$${{{u_1}} \over {{u_2}}} = {{\sqrt 3 } \over {\sqrt 2 }}$$

Both projectile reach the same maximum height.

$$\therefore$$ $${H_1} = {H_2}$$

$$ \Rightarrow {{u_1^2{{\sin }^2}{\theta _1}} \over {2g}} = {{u_2^2{{\sin }^2}{\theta _2}} \over {2g}}$$

$$ \Rightarrow u_1^2{\sin ^2}{\theta _1} = u_2^2{\sin ^2}{\theta _2}$$

$$ \Rightarrow {\left( {{{{u_1}} \over {{u_2}}}} \right)^2} = {\left( {{{\sin {\theta _2}} \over {\sin {\theta _1}}}} \right)^2}$$

$$ \Rightarrow {\left( {{{\sqrt 3 } \over {\sqrt 2 }}} \right)^2} = {\left( {{{\sin 60^\circ } \over {\sin {\theta _1}}}} \right)^2}$$

$$ \Rightarrow {3 \over 2} = {3 \over {4{{\sin }^2}{\theta _1}}}$$

$$ \Rightarrow \sin _{{\theta _1}}^2 = {1 \over 2}$$

$$ \Rightarrow \sin {\theta _1} = {1 \over {\sqrt 2 }} = \sin 45^\circ $$

$$ \Rightarrow {\theta _1} = 45^\circ $$