Given: x = -86, y = 75.
tanA = 75/-86 = -0.8721,
A = -41.09 deg.
-41.09 is in 4th quad., but our vectors are in 2nd quad. so we add 180
deg:
-41.09 + 180 = 138.9 deg.
The x component of a certain vector is -86.0 units and the y component is +75.0 units.
The magnitude of the vector is 114 m.
What is the angle between the direction of the vector and the positive direction of x? Use the convention that positive angles are measured counterclockwise from the +x axis and negative angles are measured clockwise from the +x axis.
1 answer
1 answer