For angles of projection of a projectile at angles $$\left( {{{45}^ \circ } - \theta } \right)$$   and $$\left( {{{45}^ \circ } + \theta } \right),$$   the horizontal ranges described by the projectile are in the ratio of

Solution :
We know that, horizontal ranges for complementary angles of projection will be same. The projectiles are projected at angles $$\left( {{{45}^ \circ } - \theta } \right)$$  and $$\left( {{{45}^ \circ } + \theta } \right)$$  which are complementary to each other i.e. two angles add up to give $${90^ \circ }.$$ Hence, horizontal ranges will be equal. Thus, the required ratio is $$1:1.$$
Alternative
Horizontal range of projectile $$R = \frac{{{u^2}\sin 2\alpha }}{g}$$
$$\eqalign{ & {\text{For}}\,\,\alpha = \left( {{{45}^ \circ } - \theta } \right) \cr & {R_1} = \frac{{{u^2}\sin 2\left( {{{45}^ \circ } - \theta } \right)}}{g} \cr & = \frac{{{u^2}\sin \left( {{{90}^ \circ } - 2\theta } \right)}}{g} \cr & = \frac{{{u^2}\cos 2\theta }}{g} \cr & {\text{For}}\,\,\alpha = \left( {{{45}^ \circ } + \theta } \right), \cr & {R_2} = \frac{{{u^2}\sin 2\left( {{{45}^ \circ } + \theta } \right)}}{g} \cr & = \frac{{{u^2}\sin \left( {{{90}^ \circ } + \theta } \right)}}{g} \cr & = \frac{{{u^2}\cos 2\theta }}{g} \cr & {\text{Hence,}}\,\frac{{{R_1}}}{{{R_2}}} = \frac{1}{1} \cr & {\text{or}}\,{R_1}:{R_2} = 1:1 \cr} $$