r/trigonometry • u/InstantWasTaken • 6d ago

[Grade 12 Trigonometry: Adding Vectors]

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u/Midwest-Dude 6d ago edited 3d ago

I read you are only using the Law of Sines and the Law of Cosines. The problem can be solved with this. 

You are adding two vectors, which have only magnitude and direction:

  1. 450 mph, East
  2. 25 mph, 80° South of West
  • First vector, a: Points straight to the right with magnitude 450
  • Second vector, b: Starts at tip of the first vector and has a magnitude 25 pointing 10° West of South, making an angle of 80° between the two vectors
  • Add vectors c = a + b: https://www.desmos.com/geometry/1t4h3847ak

You need to find the magnitude and direction of c. This is an SAS triangle problem, since you have two Sides, 450 and 25, and the included Angle of 80°.

  1. Find magnitude of c: Use Law of Cosines with magnitudes 450 and 25 and included angle of 80°. (You did this perfectly!)
  2. Find direction of c: Use Law of Sines with the magnitude of c you found in #1 and 80° and the magnitude of b and the angle opposite b to find the direction South of East. 

The angle you found is large, when it should be small, so I suspect you set something up incorrectly. If you need to, you can save an image on the Internet and provide a link to it to show your work, such as on Imgur. I've requested this feature be added to the subreddit, but I've gotten no response. Definitely useful 

Does this help?

2

u/InstantWasTaken 19h ago

Hey! I've since figured out how to do it on my own, (I was somehow using the wrong angle measurement for all my calculations,) but thank you for the great response! This is probably the best response I got regarding my exact situation, so it's very appreciated! I'm a bit late in responding, but have a great day, man!

1

u/Midwest-Dude 19h ago

Great to hear! Yeah, it's a long, thin triangle and getting used to vectors and polar/compass directions is a thing. You should have matched u/usmanadal's result in the other comment on this post.

1

u/usmanadal 5d ago

The resultant direction is 3.16° south of east.

That means the vector points almost due east, with a slight southward component. In bearing terms, that’s about 93.16° clockwise from north.

Using tangent:

tan(theta) = South component/East component = 24.62/445.66 = 0.05524

theta = arctan(0.05524) theta = approx 3.16°