Physics is killing me! PLZ HELP!

The projectile launched at ___ degrees would remain in the air the longest. The projectile launched at ___ degrees would have the greatest horizontal displacement. The projectile launched at ___ degrees would have the same horizontal displacement as the projectile launched at 20 degrees.
Enter your 3 answers in respective order.
a. 0
b. 5
c. 10
d. 15
e. 20
f. 25
g. 30
h. 35
i. 40
j. 45
k. 50
l. 55
m. 60
n. 65
o. 70
p. 75
q. 80
r. 85
s. 90

We know that
x = v*cos(a)*t
y = v*sin(a)*t - g*t^2
for the laws of motion.

First question: 90°
For the projectile that remains in air the longest, you want to consider that the time of the flight can be found by letting y = 0 (i.e. the projectile is at ground level).
y = v*sin(a)*t - g*t^2
0 = v*sin(a)*t - g*t^2
0 = t ( v*sin(a) - g*t )
which gives you t = 0 (which is the instant right after you shoot) or t = v*sin(a)/g (the actual flight time)
You can easily see that it is maximum when sin(a) is maximum, so when a is 90°.

Second question: 45°
you know that the displacement is x = v*cos(a)*t and that the time of the flight is v*sin(a)/g.
Put it all together to get x = v^2*sin(a)*cos(a)/g. This one hits its maximum for a = 45°. (You can easily see it by calculating its derivative)

Third question: 70°
The horizontal displacement is v^2*sin(a)*cos(a)/g. Forget about the other variables, and focus on a.
We know that sin(x) = cos(90-x) and cos(x) = sin(90-x)
thus, sin(20) = cos(70) and cos(20) = sin(70)
sin(20)cos(20) = cos(70)sin(70).