A body is acted upon by a force A of 31 Newtons directed North and also by a force B of 20 Newtons at the same point directed East. What force R in Newtons must be applied to the body to produce equilibrium?

Let the angle of force producing equilibrium be in decimal degrees measured anticlockwise from the east.

1) Find the Equilibrant in Newtons
2) Angle of Equlilbrant in decimal degrees

For number 1, I came up with 36.891N
For number 2, I came up with 49.822 decimal degrees

Is this correct?
Thank you

For #1, I get the same thing, but I rounded it off to 36.892N (36.8917).

For #2, I get 57.171 degrees from the east. Note that the angle is measured anticlockwise from the east. So, east is 0 degrees. If R is the resultant vector with magnitude 36.892 N and θ is the angle that it makes with the zero degrees (the horizontal on a Cartesian coordinate system),

R \,cos(\theta) = 36.892\, cos(\theta) = 20

cos(\theta) = \frac {20}{36.892} = 0.542

\theta = arccos(0.542) = 57.172\,degrees

You can also verify that

R\,sin(\theta) = 36.892\,sin(\theta) = 31

I would like to point out that it is preferable to work with the values that you were given, that is the 31 N and 20 N. The reason for this is to reduce the possibility of an 'error carried forward'. Say you wrongly pressed the buttons on your calculator and you were not aware of it, then passed on to the next part. If you used the value you just got, you are bound to include a mistake, making your answer to the second part wrong. That's why, here, you'd prefer to use the tan inverse of 31/20, giving 57.1714 = 57.2 degrees.

For the first part, give your answer to 2 significant figures, for the highest significant figure is 2 in the whole question. For angles, one decimal place is generally required.

Thanks for your help, guys. I really appreciate it.