Physics HW*vector* problem?

Question

The Vector V1 is 8.08 units long and points along the negative X axis.The Vector V2 is 4.51 units long and points at +45.0* to the positive X axis.
1.What are the X and Y components of each vector?
2.Determine the sum of two vectors(magnitude and angel)

Answer 1

V1 is -8.08x+0y
V2 is 4.51cos(45)x+4.51sin(45)y = 3.189x+3.189y

Sum is -4.891x+3.189y
mag is square root of sum of squares = 5.8388
direction is tan inverse of y over x = -33.1 degrees

Answer 2

The X and Y components of vector V1 are

V1_x = (8.08 units)*cos(180 degrees) = -8.08 units
V1_y = (8.08 units)*sin(180 degrees) = 0 units

The X and Y components of vector V2 are

V2_x = (4.51 units)*cos(45 degrees) = 3.189 units
V2_y = (4.51 units)*sin(45 degrees) = 3.189 units

To sum them, add the magnitudes:

sum_x = (-8.08 + 3.189) units = -4.891 units
sum_y = (0 + 3.189) units = 3.189 units

The magntude of this vector is:

|sum| = sqrt( 4.891^2 + 3.189^2 ) = 5.839 unts

The angle of this vector is arctan( -3.189/4.891 ) = -33.105 degrees, but you have to interpret this answer in the SECOND QUADRANT. That is,

angle( sum ) = 33.105 degrees "north" of "west" (clockwise from horizontal)

or

angle( sum ) = (180 - 33.105) degrees = 146.895 degrees (counter-clockwise from horizontal)

Answer 3

(1)
component along the X axis,
X= 8.08 cos 180 + 4.51 cos 45 = -4.8909
Y= 8.08 sin 180 + 4.51 sin 45 = 3.1891

(2)
these 2 concurrent vectors have an angle of 135 between their lines of action.
Resultant,
√{(8.08)^2+(4.51)^2+2*8.08*4.51*cos135}
=10.222

if it works in Ө angle with V1, then

tan Ө =(4.51 sin 135)/(8.08+4.51 cos 135)=0.652
Ө=33.11

thus the resultant will act in (180-33.11)=146.89 degree having a value of 10.222 units