Of course the angles are measured with your protractor. No scale is required for that measurement.
You can purchase a protractor from the bookstore or use one from the  laboratory. 

Note theta is the angle with the x- axis.

Now  use your basic trigonometry skills to find both components with respect to the  other angle.  Use the properties of the sine and cosine of complimentary angles. The second angle, with the y-axis,  is the compliment of the first because they add to  90 degrees.

(a) The y - component of this vector is zero. The x component is  -v·cos0.
(b) Points in Quadrant 3 and  makes angle 40 degrees with the negative y-axis. Use your knowledge of complimentary angles to find  angle theta  with the negative x-axis.
 Both vector components  are negative: Thus a_x = -a·cos(theta) and a_y = -a·sin(theta) . Recall  a is the magnitude.
(c) Points in Quadrant 1 and  makes angle  36.9 degrees with the positive y-axis. Use your understanding of complimentary angles to determine  angle theta  with the positive x-axis.
Both vector components  are positive: Thus F_x = F·cos(theta) and F_y = F·sin(theta) . Recall  F is the force's magnitude. We will learn about the force vector  in Chapter 5. 

Manually: 

The vector,  relative to  reference frame (a),   is (-Bcos30, Bsin30) = (B_x , B_y).  Draw  frame (b) on top of frame (a). The two axes must share the same origin. Let the origin coincide with  vector-B's tail.

Accurately draw the x'-axis rotated clockwise  from the x-axis by 30 degrees. Likewise sketch carefully the y'-axis  rotated from the  y-axis by the same angle. Then drop  perpendiculars from vector-B's tip  to the x' and y'-axe. Now identify the components  in the  rotated  frame.  

Classic formula:

Anybody with persistence and trig.  exposure  can  derive the following relationship between coordinates in the two systems.

x' = xcos30 + ysin30

y' = -xsin30 + ycos30

This is true for ANY vector, for example vector-B:

B_x' = B_xcos30 + B_ysin30

B_y' = -B_xsin30 + B_ycos30

If you want, I can  geometrically derive these relations  in lecture or office hours. Stay tuned. 
-------------------------------------------------------------------------------------------------------------------------------- 
You can easily show  the magnitude of any vector is unchanged under rotations:

x'²  + y'² = x² + y²   

and

B_x'² + -B_y'²   =  B_x² + B_y² 

(a) Here's a hint to get you started on the other parts. Vector -B points in Quadrant 2. 
 B_x = -4 and B_y= 4. Find the magnitude using the Pythagorean Theorem. The reference angle with the negative x-axis
 is  tan^-1(4/4) = the angle whose tangent is 1.  We all know what the angle is, so I'll stop here.
Below,  find the magnitude and related angle in addition to the Quadrant: 
(b) Vector points in Quadrant 3. 
(c) Quadrant 3. 
(d) Quadrant 1.  

Note this EXPLANATION makes more sense for section 02, which dealt with all 
this last WED with the handouts.
Section 01's experiment is in the future. I can provide the lab handouts 
during the  next lecture.
Either way, this message is useful.

There are two logical choices based upon the handout on graphing by 
hand. Below applies to the
fresh graph paper I gave you.

1) For the vertical axis ( for v in cm/s),
5 darkest major divisions (i.e. 50 small divisions ) represents an 
interval of 100 units (100 cm/s). Thus 1
small division represents an interval of 2 units (2 cm/s) .

This scaling could work for your data with smart choices for where 100 
cm/s is marked on the vertical axis.
To capture most of the data points you can:

(i) set the 100 cm/s mark to be ten small divisions from the bottom of 
the vertical axis, 
(ii) count 50 small divisions up (count 5 darkest major divisions up) to 
mark 200 cm/s,
(iii ) then count count 50 more small divisions up (count 5 more darkest 
major divisions up) to mark 300 cm/s,
(iv) and finally  count 30 small divisions up ( count 3 more darkest 
major divisions up) to mark 360 cm/s.

In this scenario, 360 cm/s will be marked at the very top of the 
vertical axis.

Please make appropriate adjustments for marking the first  velocity 
value  on the vertical axis
to make sure the extrapolated "best fit" line goes through the vertical 
axis
at t = 0. You want to be able to see the initial velocity ( at t = 0.)

In this scenario, you could choose  5 small divisions to represent a 
time interval of 1/60 sec on the horizontal axis.

2)  This next choice has poorer resolution.
For the vertical axis ( for v in cm/s),
5 alternating darkest and less dark major divisions (i.e. 25 small 
divisions ) represents an interval of 100 units 100 cm/s). Thus 1
small division represents an interval of  4 units (4 cm/s) .

This scaling would work for your data, without the worry that not all of 
the data points will fit on the graph paper, but with
a loss of resolution in the graphing process. To capture most of the 
data points you can:

(i) set the 100 cm/s mark to be 25 small divisions from the bottom of 
the vertical axis, 
(ii) count 25 small divisions up (count 5 alternating darkest and less 
dark major divisions up ) to mark 200 cm/s,
(iii ) then count  25 more small divisions up (count 5 alternating 
darkest and less dark major divisions up) to mark 300 cm/s,
(iv) and finally  count  25 more small divisions up (count 5 alternating 
darkest and less dark major divisions up) to mark 400 cm/s, etc


The extrapolated "best fit" line is guaranteed to go through the 
vertical axis
at t = 0 such than you can  see the initial velocity ( at t = 0.)

In this scenario, you could also choose  5 small divisions to represent 
a time interval of 1/60 sec on the horizontal axis..

SUMMARY  FROM THE HANDOUT

Yo, what do these two above scenarios have in common in conjunction with 
the provided graph paper that works well with the decimal system? 
Read the handout again to find out: The bottom line is ( see page 59 top 
and Fig. 5.1) that the number of units between adjacent major divisions
must be 1x10^N, 2x10^N or 5x10^N.
# In scenario (1) above, on the vertical axis we had 1x10^N with N = 2, 
which equals 100 units between two adjacent darkest
major divisions (10 small divisions apart).
# In scenario (2) above, on the vertical axis we also had 1x10^N with N 
= 2, which equals 100 units between alternating darkest and
less dark adjacent major divisions (5 small divisions apart.)
# In both scenarios, on the horizontal  axis, we had 5x10^N with N = 0, 
which equals  the number of units between adjacent darkest
and less dark major divisions (separated by 5 small divisions.). Note: 
One unit = 1/60 seconds. 5 small divisions = 1/60 second.

Finally, in Figure 5.1, also explained on page 59 top, on the vertical 
axis, 1 darkest major division (i.e. 10 small divisions ) represents an 
interval of 20 units.
That's 2x10^N with N = 1. Thus 1 small division represents an interval 
of 2 units. On  the horizontal axis, 1  darkest major division (10 small 
divisions)
represents 5 units. That's 5x10^N with N = 0. Note that  we consider the 
time values in all cases to be exact.

Remember that all of this discussion assumes the decimal based graph 
paper  provided free of charge in the lab sessions.