Calculating Resultant Force Magnitude and Direction Angle

[freespirit]

My problem is to determine the magnitude and direction of the resultant force FR=F1+F2 and it's direction, measured counterclockwise from the positive x direction.

f1=250 lb @ 60 degrees from x
f2= 375 lb @ -45 degrees from x

Ok I got the magnitude by doing this:
(360-2(255))/2=-75 degrees

fr=sqroot of (250^2+375^2-2(250)(375)cos(75)
fr=393.188~ 393

then I got the angle by this:
375/sin x = 393.188/sin 75
x=67.1088
how do i get the resultant angle, what do I need to add to the 67 degrees?

 

[nautica]

breakdown x and y components of each by the following

X= r cos(angle)
Y= r sin(angle)

Add the two x and y compents to get the x and y of the resultant. Use the pathagorean theroum to get the resultant magnitude and use tan^(-1) (y/x) to get resultant directions.

nautica

 

[himanshu121]

Preview Rough fig in attachment so

$$tan{\alpha} = \frac{ysin{\theta}}{x+ycos{\theta}}$$

$$tan{\alpha} = \frac{ysin{\theta}}{x+ycos{\theta}}$$

 

[freespirit]

Thank You

Thank you both for your help.