Solve the Problem $$\mathrm { r } = \frac { 1 } { 32 } \mathrm { v } ^ { 2 } \sin 2 \theta ,$$

Solve the problem.
-The range r of a projectile is given by $$\mathrm { r } = \frac { 1 } { 32 } \mathrm { v } ^ { 2 } \sin 2 \theta ,$$
where $\mathrm { v }$ is the initial velocity and $\theta$ is the angle of elevation. If $r$ is to be $3000 \mathrm { ft }$ and $\mathrm { v } = 2000 \mathrm { ft } / \mathrm { sec }$ , what must the angle of elevation be? Give your answer in degrees to the nearest hundredth.

A) $89.31 ^ { \circ }$
B) $1.38 ^ { \circ }$
C) $0.96 ^ { \circ }$
D) $0.69 ^ { \circ }$