Calculate Angle C with Tail-Tip Method

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Homework Help Overview

The discussion revolves around calculating angle C in a triangle formed by two vectors using the tail-tip method. The original poster has two vectors but lacks the resultant vector's length and the angles associated with them.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of the law of cosines and the law of sines, questioning the original poster's understanding of triangle notation and the information available. There is a focus on the need for the length of the resultant vector and the angles to proceed with calculations.

Discussion Status

Some participants have offered guidance on using vector components to find the resultant vector's magnitude, while others emphasize the need for additional information to solve for angle C. The conversation reflects a mix of interpretations regarding the available data and the methods to apply.

Contextual Notes

The original poster has indicated they do not have the length of the resultant vector or the angles of the vectors, which raises questions about the feasibility of solving for angle C with the given information.

Confused one
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I have 2 vectors, which are plotted using the tail-tip method to form a non-right triangle.

Side A is vector 1, side B is vector 2, and side C is my resultant.
I know I can get angle C, but that's where I'm stuck.

How do you get angle C? In simple terms...:confused:
 
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It's a bit confusing to talk about "side C" and "angle C"! Standard notation is that the lengths of the three sides of the triangle are a, b, c and the angles opposite each are A, B, C respectively. If you know a, b, c, then use the cosine law to find C: c2= a2[/sup]+ b2- 2ab cos(C). Plug in a, b, c and solve for C.
 
I don't have c. I have a, b, and their degrees, but no c. I don't have A or B either.

I'm trying to sove for C, but I must have c before I can use the law of cosines.
 
how about law of sins

sin A / a = sin B / b = sin C / c
 
You orginally said "Side A is vector 1, side B is vector 2, and side C is my resultant" so it was reasonable to assume that you knew those 3 lengths. In vector problems, normally you know two vectors, say lengths a and b, and the angle between them, but that would be C which you say you don't know. If lengths a and b are the only values you have, you can't solve this: two parts of a triangle are not enough! What information do you have?
 
You have to understand that the resultant vector components are the sum of the adding vectors. Add the vertical component of the vector A to the vertical component of vector B. Then add the horizontal component of the vector A to horizontal component of vector B. Then you have the components of vector C and the magnitude of C = ((Ay + By)^2 + (Ax + Bx)^2)^1/2
 
Werg22 said:
You have to understand that the resultant vector components are the sum of the adding vectors. Add the vertical component of the vector A to the vertical component of vector B. Then add the horizontal component of the vector A to horizontal component of vector B. Then you have the components of vector C and the magnitude of C = ((Ay + By)^2 + (Ax + Bx)^2)^1/2

That's one way to do it. It may or may not be simpler than adding the vectors "geometrically". The question "Confused one" initially asked made it clear, I thought, that he was attempting to do this geometrically. I sure wish he would get back to us and tell us exactly what information he has to work with. I would have suggested that he find angle C by just subtracting the angles the the vectors make with some fixed line, but surely he wouldn't be having so much trouble if he were given the vectors like that.
 

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