(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Find y component of vector C from its length and the angle it makes with the x axis, that is, from geometry. Express the y component of vector C in terms of C and [itex]\phi[/itex].

2. Relevant equations

Vector addition using geometry:

1) C = [itex]\sqrt{A^{2}+B^{2}-2ABcos(c)}[/itex]. Law of Cosines

2) [itex]\phi[/itex] = sin[itex]^{-1}[/itex][itex]\frac{Bsin(c)}{C}[/itex]. Law of Sines

Vector addition using components:

3) C[itex]_{x}[/itex] = A + Bcos[itex]\vartheta[/itex].

4) C[itex]_{y}[/itex] = Bsin[itex]\vartheta[/itex]

3. The attempt at a solution

Since I'm supposed to use geometry, I'm limited to using equations 1 or 2 (Law of cosine and sin)

I could easily determine C[itex]_{y}[/itex] by making a right triangle and forming the equation: [sin[itex]\phi[/itex]=[itex]\frac{Cy}{C}[/itex]] = C[itex]_{y}[/itex] = Csin[itex]\phi[/itex]

However, that would be using the component addition method. I don't see how I would use geometry (Law of sine or cosine) to produce the same answer of Csin[itex]\phi[/itex].

The only thing I could come up with was taking SAS (side angle side) [itex]\sqrt{C}[/itex][itex]_{y}[/itex][itex]^{2}[/itex] = [itex]\sqrt{C^{2}+C_{x}^{2}-2C_{x}Ccos\phi}[/itex] , but then how does that equate to

C[itex]_{y}[/itex] = Csin[itex]\phi[/itex] ?

Am I supposed to find the side using the Law of Cosine equation? If so, then

Have I described the correct way of proceeding with the problem? If so, then

Did I come up with the correct Law of Cosine forumula? If so, then

Am I supposed to put it in the same form that I would have gotten using component addition? ( C[itex]_{y}[/itex] = Csin[itex]\phi[/itex]) ? If so, then

How does [itex]\sqrt{C^{2}+C_{x}^{2}-2C_{x}Ccos\phi}[/itex] turn into Csin[itex]\phi[/itex] if they are essentially equal to each other and represent C[itex]_{y}[/itex]?

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# H/W help on Geometric vs Component Vector Addition

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