How do I solve this vector problem involving three ropes and a rough surface?

[y90x]

Homework Statement

I’m working on a problem that requires adding 3 vectors . And while I was following the book, it said to find the angle to divide the y over the x components of the resistant , and showed the angle to be -1.25 but then follow d to change to -51 .
How did it change ? The book doesn’t explain I’ll inclide a photo of the textbook so you can see what I’m talking about
Here’s the problem I’m working on just in case:

Three horizontal ropes are tied to a 30 kg mass resting on a rough surface. The coefficient of friction between the mass and the surface is 0.12. Rope A has tension 60 N directed Northwest, rope B has tension 70 N directed 20° East of South, and rope C has tension 80 N directed 35 North of West. Find the resulting acceleration of the 30 kg mass.

Homework Equations

The component method for adding vectors

The Attempt at a Solution

Rope A :
-60cos(45) + 60sin(45)= A
-42.426 + 42.426 = A
Rope B :
70cos(20) + -70sin(20) = B
65.778 + -23.94
Rope C:
-80cos(35) + 80sin(35) = C
-65.532 + 45.886 = C

To find the resultant magnitude :
Square root of ( all the X components added up) (-42,1800)^2 + 64.3711^2 ( all the Y components added up)
= 76.959

To find the angle :
Pheta = tan^(-1)(Y/X)
= tan^(-1) (64.3711/-42.1800)
= -56.8 Solution of the actual problem :
1.72 m/s^2 at 15.0 north of westhttps://www.physicsforums.com/attachments/215616

 

[TSny]

rope B has tension 70 N directed 20° East of South,

The Attempt at a Solution

Rope B :
70cos(20) + -70sin(20) = B
65.778 + -23.94

Note that 20° East of South is not the same as 20° South of East

Also, do you need to add in the friction force?

 

[y90x]

Note that 20° East of South is not the same as 20° South of East

Also, do you need to add in the friction force?

You’re right , thanks