First, planes fly on headings, not bearings.
Second, winds are usually indicated by where they are blowing from, but I'll assume your wind is indeed blowing roughly westward.
285° is measured clockwise from due north, which is 90°. So, it is in QII. Due west is a heading of 270°, so it is 15° north of west.
480@285° = <480cos285°,480sin285°> = <-480cos15°,480sin15°>
30@265° = <30cos265°,30sin265°> = <-30cos5°,-30sin5°>
To find the actual speed and direction of the plane, convert the values to decimal values, add the components, and then use the Pythagorean theorem to find the speed, and then figure the angle (tanθ=y/x). That θ is measured counterclockwise from the x-axis, so you need to convert that to actual course heading.
An airplane is flying on a bearing of 285 degrees at 480 mph. A wind is blowing with the bearing 265 degrees at 30?
a. find the component from of the velocity of the airplane
b. find the actual speed and direction of the airplane
485 sin 75 degrees
-485 cos 75 degrees (in 4 th quadrant)
30 sin -20 degrees
-30 cos -20 degrees
Then, added them for component form.
Here is what I got, but it is not correct....I'm really not sure how to start the problem and finish it with the actual speed and direction.
1 answer
1 answer