How Can You Calculate the Magnitude of Vector A Using Dot Product Rules?

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The discussion revolves around calculating the magnitude of vector A using dot product rules in a physics problem. Given vectors B and C with specific angles and magnitudes, the user derives two equations involving cosine functions. They attempt to simplify these equations by dividing them but encounter difficulties with the expansion of the cosine function. Ultimately, the user resolves their confusion and successfully figures out the solution. The thread highlights the importance of understanding trigonometric identities in vector calculations.
rambo5330
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Homework Statement


i'm having troubles solving this problem in physics involving dot product rule


Vector B is 5m and 60 degress above the x axis
A has the same magnitude as C and C is has 25 more degrees than A ..
find magnitude of A..
B dot C = 35
B dot A = 30

this all we're given.. as far as i can get it i make it down to two equations
cos(\theta + 25) = 7/C and cos(\theta) = 6/C
 
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The two equations should be

cos(θ-60 + 25) = 7/c...(1)
cos(θ-60) = 6/c ...(2)
divide (1) by (2), you get

\frac{cos(\theta-60 + 25))}{cos(\theta-60)} = 7/6

Let (θ - 60) = x
expand cos(x + 25) and divide it by cos(x)and simplify to find x.
 
excellent thank you,

I thought that was the method but I'm still running into trouble expanding cos(x + 25) and dividing to clear... when i expand i get


cosx / (cos(x))(cos(25)) - (sin(x))(sin(25)) = 6/7

i don't see how dividing that by cos(x) now accomplishes anything?

where am i going wrong?
 
nm its all good i got it figured out thanks !
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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