Question

Let the angle θθ be the angle that the vector A⃗A→makes with the +x-axis, measured counterclockwise from that axis. Find the angle θθfor a vector that has the following components.

PART 1:

Ax= 3.00 mm , Ay= -0.500 mm

Express your answer in degrees.

PART 2:

Ax= 1.80 mm , Ay= 3.30 mm

Express your answer in degrees.

PART 3:

Ax= -3.00 mm , Ay= 3.20 mm

Express your answer in degrees.

PART 4:

Ax= -1.20 mm , Ay= -3.40 mm

Express your answer in degrees.

Answer 1

Suppose given that Vector is R and it makes angle with +x-axis, then it's components are given by:

Rx = Horizontal component = R*cos

Ry = Vertical component = R*sin

And direction of vector will be given by, = 2n*pi +/- arctan (Ry/Rx)

Using above rule:

Part 1.

Ax = 3.00 & Ay = -0.500, (Since Ax > 0 and Ay < 0, So Vector A is in 4th quadrant)

So,

= arctan (-0.500/3.00) = 360 deg - arctan (0.500/3.00) = 360 - 9.46

= 350.54 deg

Part 2.

Ax = 1.80 & Ay = 3.30, (Since Ax > 0 and Ay > 0, So Vector A is in 1st quadrant)

So,

= arctan (3.30/1.80)

= 61.39 deg

Part 3.

Ax = -3.00 & Ay = 3.20, (Since Ax < 0 and Ay > 0, So Vector A is in 2nd quadrant)

So,

= arctan (3.20/(-3.00)) = 180 - arctan (3.20/3.00) = 180 - 46.85

= 133.15 deg

Part 4.

Ax = -1.20 & Ay = -3.40, (Since Ax < 0 and Ay < 0, So Vector A is in 3rd quadrant)

So,

= arctan ((-3.40)/(-1.20)) = 180 + arctan (3.40/1.20) = 180 + 70.56

= 250.56 deg

Let me know if you've any query.