Finding the angle between two magnitudes.

[Daweih]

Consider two displacements, one of magnitude 3.3 m and another of magnitude 4.3 m. What angle between the directions of this two displacements give a resultant displacement of magnitude (a) 5.1 m, (b) 2.8 m, and (c) 4.7 m.

Homework Equations

I've been using the law of cosine to work out this problem, but my answers have all come out to be wrong. Am I incorrect in using it to solve this problem or am I just doing my math wrong?

c^2 = a^2 + b^2 - 2abcosγ


My answers are as follows: a) 83° b) 41° c)75°

I would really appreciate the help. =(

 

[SammyS]

Consider two displacements, one of magnitude 3.3 m and another of magnitude 4.3 m. What angle between the directions of this two displacements give a resultant displacement of magnitude (a) 5.1 m, (b) 2.8 m, and (c) 4.7 m.

Homework Equations

I've been using the law of cosine to work out this problem, but my answers have all come out to be wrong. Am I incorrect in using it to solve this problem or am I just doing my math wrong?

c^2 = a^2 + b^2 - 2abcosγ

My answers are as follows: a) 83° b) 41° c)75°

I would really appreciate the help. =(

Without working the problem myself: (Don't you love it when someone qualifies their answer this way?)

I suspect that your answers are correct for the angle the vectors make if you lay them head to tail, as you would in constructing a triangle composed of the two vectors and their resultant.

However, that is not the same as the angle between the vectors directions. The angle between the vectors directions is the supplement of the angle between the vectors when placed head to tail.

 

[Daweih]

Ah, I understand now. Thank you very much. That clears up everything for me. =)

 

[FeynmanIsCool]

You are correct with Law of Cosines. I got 83.18 for a, so your data is correct. How are you drawing a,b and c? Hopefully you can see that (for a) C=5.1m
I can't just give you the answer, but try a variation of law of cosines:

C=cos^-1(a²+b²-c²)/(2ab)

Try that.