I Keep Getting 2 Different Answers.

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The discussion revolves around calculating the resulting velocity of a plane flying north at 200 km/h with a northeast wind of 100 km/h. Participants are confused about obtaining two different angles for the same resultant speed of 147.36 km/h, specifically 118.07 degrees and 75.361 degrees. Various methods are suggested for vector addition, with emphasis on drawing accurate diagrams to visualize the vectors involved. The importance of using the correct angle measurements and applying trigonometric formulas is highlighted, with some users suggesting the cosine rule for clarity. Ultimately, the conversation seeks to confirm the correct angle and resultant speed for the plane's velocity relative to the ground.
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I Keep Getting 2 Different Answers. Please Help!

1. Plane has a speed of 200 km/h and heads due north; though, 100 km/h northeast wind blows. what is the resulting velocity of plane w/ respect to ground?



2. I keep getting two different answers (I've approached them 2 different ways). I get 147.36 km/h both times but the angle is different! I either get 118.07 degrees and 75.361 degrees. Can anyone confirm one of them?



3. TO get 118.67, I drew the vectors of (-70.71 and 129.29 km/h ---> also tan(x) = 129.29/70.71 ...For the other angle of 75.361, i had tan(x) = 270.711/70.711... BASICALLY I KEEP DRAWING IT DIFFERENTLY :/ Can anyone tell me which ones right?
 
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your this thing "129.29 km/h" seems wrong to me.

I get 270.711

anywyas, how you got 129?
 
thank you! can anyone else confirm?
 
I don't understand where your value of -70.71 comes from? Draw your diagram so you have the 200 km/h in a north direction, and the 100km/h in a southwest direction at the end of the north arrow (a northeasterly wind means a wind blowing from the northeast). Then, the other line in this triangle is the resultant. You can find the angle using the cosine rule for triangles. (NB: the actual angle you want will be the angle of the vector from due north measured in a clockwise way)
 
rootX said:
your this thing "129.29 km/h" seems wrong to me.

I get 270.711

anywyas, how you got 129?

Exactly, i am getting 279.8008km/hr. Take one vector as the plane's velocity and the other vector as the wind's velocity. Now the resultant of this two vectors would be the velocity of the plane w.r.t the ground. And now you can apply the formula as tanx=bsin(theta)/a+bcos(theta).
Here theta is the angle between the two vectors which is 45 and x is the angle which the resultant vector makes with the vector a. Here vector b is the velocity of the plane and vector a is that of wind. Evaluating we get tanx=0.5857. So x is 30.36 this is the angle between the resultant and a. Now a makes an angle of 45 with the positive x axis.So the angle made by the resulatant with the positive x-axis is 75.36
I think that this should be the final answer
 
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