Angle of Projection from Height and Range

[atmega-ist]

Homework Statement

A projectile is fired in such a way that that its horizontal range is three times its maximum height. What is the angle of projection?

Homework Equations

R = V_o²sin(2theta)/g

H = (V_osin(theta)²/2g

R = 3H

Cancel V_o^o, g and sin(theta) to leave 4/3 tan(theta) = 1

The Attempt at a Solution

This is, as I've seen, not a new question. I'm good with how to solve it but where I see it should be tan(theta) = 4/3, I keep getting (as stated above) 4/3tan(theta) = 1 where if I were to bring the 4/3 to the other side I would end up with tan(theta) = 3/4.

My only problem is that I can't figure out how this 4/3 is ending up on the opposite side of the tan(theta). I don't know if I'm completely missing something or if it's just because it's 1:30 am...

Thanks

 

[atmega-ist]

aaaaaaaaaaaaaaaaaaaaaaand now I feel like a REAL genius...

tan is sin/cos... NOT cos/sin...

long time since trig + reading too quickly + no sleep = embarrassment by simple math (and negative signs)

sorry