Determining the correct angle

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In summary, the plane's new velocity with respect to ground in standard location is 239.27 km/h heading 190.89 degrees clockwise from North. This was calculated by using Pythagoras' theorem to find the resultant speed and then using the cosine ratio to determine the angle with respect to South, and adding 180 degrees to get the standard location heading.
  • #1
Example question:
A plane is flying due west at 235 km/h and encounters a wind from the north at 45 km/h. What is the plane's new velocity with respect to ground in standard location?

Plane Equation (magnitude) (COS Angle) = x and (magnitude) (SIN Angle) =y
Wind Equation (magnitude) (COS ANgle) =x and (magnitude) (SIN Angle) =y
adding your x's and y's

Angle tan-1 y/x

plane angle + wind angle = adjusted angle
adjusted angle - new angle = current angle

Magnitude = square root of x squared and y squared

My attempt:
Before I start I just want to apologise for spelling all my equations out because I have no idea what I am doing here. This is my first time taking a physics course and I am completely lost

THe first issue that I was presented in this problem is finding out the angles. Which I am still unsure about. I know for example, northeast with no angle is a 45 degree angle, but if it is due north, due east, due south and due west i really have no idea. If someone can clarify that for me that would be great.

(235 km/h) (COS 0) =235 (235km/h) (SIN 0) =0
* I picked 0 for the angle (due west) because flying in a horizontial line would not have an angle?!*

(45 km/ h) (COS 90) = 0 (45 km/h) (SIN 90) =45
* I picked 90 for the angle (due north) because flying in a vertical line would create a 90 degree?!*

Current Angle:

angle tan-1 = 235/45 =79.11


Current Magnitude:

square root (235) squared + (45) squared = 239.27

New velocity to respect of ground 239.27 km/h < 190.89 WN

I tried to make an educated guess on my vector angles, but am unsure if I chose wisely. Besides checking my work can someone clarify that due north and south will always be 90 degrees and due east and west will always be 0 degrees (unless otherwise specified).

Thank you,
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  • #2
Your answer for speed is correct.

Although you could have simply used Pythagoras: A2 + B2 = C2.

2352 + 452 = C2
Sqrt(2352 + 452) = C

So now you have the resultant speed which is 239.27 and you know that will be between West and South (wind coming from the North = heading South).

To get the resultant angle, do it wrt South. So your hypotenuse is 239.27 and your adjacent is 45. Combine that with SOH CAH TOA: Cos(theta) = a/h ad that gives you the angle the aircraft will be flying in wrt South.

Now, it asks for standard location so I'd assume that means you need to give it wrt North - in this case, just add 180 (angle from North to South) and you have a heading clockwise from North which is how headings are given on an aircraft.

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