Hello, nugiboy!
Your work is correct . . . Nice going!
Did you notice that their answer is the complement of yours?
A bearing is measured clockwise from North.
Since your angle (73.7°) is measured CCW from East,
. . the bearing is: .
Im a bit stuck on this question.
- A force F is given by F = (3.5i+12j) N, where i and j are horizontal unit vectors east and north
respectively.
Calculate the magnitude of F and also its direction as a bearing.
I found the magnitude easily by using pythagerous. I can also find the angle by using the tan rule. I get an angle of 73.7', which is right. Im just not sure about putting it as a bearing. My angle is taken from the the east horizontal or x axis if you want. Eg: 0' is just a line pointing east.
Anyway could someone explain to me how to put it as a bearing. The answer should be 16.3'.
Hi thanks for the reply. I can see how they both answers fit together now. Just for future reference, is it the same with any angle if i want to change it to a bearing. By this i mean do you always minus the angle from 90'? If the angle is bigger than 90', wouldn't i get a negative bearing?
Hello, nugiboy!
Bearing is always measured clockwise from North.Do you always minus the angle from 90°? . . . . no
. . Just remember that . . .
If your angle is +120° from the positive x-axis,
. . it is in Quadrant 2.Code:\ \ ← ← \ \ \120° ↑ + - - - - -
What is the angle when measured CW from North?Code:| \ | \ | \ | /\| \ ↑ + ↓ \ / ←
Yes, you have to do some Thinking ... and some Math.
. . Sorry about that . . .
For future reference, it would not be wise to bump up a thread, especially 30 minutes after your last post. Why is there such a rush anyway? Is there a fire that I don't know about?
To answer your question, you treat the angle as if you're trying to find reference angles in Trigonometry. If it's bigger than 90 degrees, there are several other formulas you can use to find the angle measurement.