## Algebra 2 (1st Edition)

We know that $d=\dfrac{v^2}{32} \sin (2 \theta)$ $\dfrac{v^2}{9} [\sin 2 (45-k)^{\circ}]= \dfrac{v^2}{9} [\sin 2 (45+k)^{\circ}]$ or, $\dfrac{v^2}{9} \sin (90-2k)^{\circ}= \dfrac{v^2}{9} \sin (90+2k)^{\circ}$ or, $\dfrac{v^2}{9} \cos (2k)^{\circ}= \dfrac{v^2}{9} \cos (2k)^{\circ}$ Hence, the horizontal distance when $\theta= (45-k)^{\circ}$ with distance when $\theta= (45+k)^{\circ}$ is the same.