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

The correct option is A $1:1$

$\text{Step 1: Range of Projectile [Refer Fig.]}$

Let $u$ be the initial velocity of the projectile.

Range of a projectile is given by:

$R=\frac{u^2\sin 2\theta }{g}$

$\text{Step 2: Ratio of}{R}_1\&R_2$

At ${\theta }_1={45}^o-\theta ,R=R_1$

$R_1=\frac{u^2\sin 2({45}^o-\theta )}{g}$$=\frac{u^2\sin ({90}^o-2\theta )}{g}$

At ${\theta }_2={45}^o+\theta R=R_2$

$R_2=\frac{u^2\sin 2({45}^o+\theta )}{g}$$=\frac{u^2\sin ({90}^o+2\theta )}{g}$

The ratio of $R_1$ and $R_2$ is given by

$\frac{R_1}{R_2}=\frac{\sin ({90}^o-2\theta )}{\sin ({90}^o+2\theta )}=\frac{\cos 2\theta }{\cos 2\theta }=1$

Hence the ratio of the range of the projectile is $1:1$. Option $A$ is correct.