The water from a fire hose exerts a force of 200 lbs on the person holding the hose. The nozzle of the hose weighs 20 lbs.What force is required to hold the hose horizontal? At what angle to the horizontal is the force applied?I drew a graph on a x,y coordinate system

I have the Force hose on the -x axis (-200,0) then from the end of that point i have the point (0,-20) [or it would be (-200,-20) from (0,0)] for the weight of the hose.

so i just thought about it and i said to counter act the force of the hose and the weight you would draw a triangle with the points (200,0) then from the end of that point (0,20).

The hypotenuse of this triangle would be $\displaystyle \sqrt(40400) $

So would the force required to hold the hose horizontal would be $\displaystyle \sqrt(40400) lbs$ ??

then then angle to the horizontal would be

$\displaystyle tan^{-1}(\frac{20}{200}) = 5.71 degrees $

Thanks for any help