Solve the Problem $\mathrm { r }$ Of a Projectile Is Given By

Solve the problem.
-The range $\mathrm { r }$ of a projectile is given by
$$\mathrm { r } = \frac { 1 } { 32 } \mathrm { v } ^ { 2 } \sin 2 \theta \text {, }$$
where $\mathrm { v }$ is the initial velocity and $\theta$ is the angle of elevation. If $\mathrm { r }$ is to be $8000 \mathrm { ft }$ and $\mathrm { v } = 3000 \mathrm { ft } / \mathrm { sec }$ , what must the angle of elevation be? Give your answer in degrees to the nearest hundredth.

A) $1.14 ^ { \circ }$
B) $1.63 ^ { \circ }$
C) $89.19 ^ { \circ }$
D) $0.81 ^ { \circ }$