knowing the following:
standard vector notation:
46.9 km@ 32 degrees
magnitude: 46.9 km
theta 58
quadrant 1
x component 24.9
y component 39.7
What would the navigational angle be
Navigation angle is measured from N, your angle is measured from E.
Navigation: 90-32=58
To determine the navigational angle, we need to consider the quadrant in which the vector lies.
In this case, since the given angle is 32 degrees and the quadrant is 1, the navigational angle would be equal to the given angle's complement to 90 degrees.
To find the complement of an angle, subtract it from 90 degrees.
Complement of angle = 90 degrees - Given angle
Complement of 32 degrees = 90 degrees - 32 degrees = 58 degrees
Therefore, the navigational angle is 58 degrees.