Projectile Motion Trig Problem

[Phyzwizz]

I found one of the answers to the problem, and I took a peak, but couldn't find out how the book got the second answer.

Problem-
A batted baseball leaves the bat at an angle of θ with the horizontal and an initial velocity of v₀=100 feet per second. The ball is caught by an outfielder 300 feet from home plate. Find θ if the range r of a projectile is given by.

r=1/32v₀²sin2θ

so 300=(1/32)(100)²sin2θ

θ=theta

I got approximately 37 degrees for one of my answers. I tried finding the second angle measure by taking the square root of 100 so I could have 2 answers but I realized that would just give me -37 degrees. I'm sure when someone answers this I will be pretty upset that I didn't see how to get the second degree angle, which the book says is 53 degrees. I plugged it into the equation and it works but how in the world do you get there.

 

[rl.bhat]

Suppose you got sin(2θ) = 1/2, then 2θ = 30 degrees or ( 180 - 30 ) degrees.
Hence you get two values. In general, in projectile motion, range will be the same for complementary angles ( whose sum is 90 degrees).