That's what I did.
Rearrange the first equation
A=sqrt(2^2+5^2-2(2)(5)cosθ)
sinφ/5 = sinθ/A to isolate sin(phi)
φ=arcsin[(5sin(θ)/A)]
should there be a parenthesis between 5 and sin?
plug that into F/W = cos(theta)/sin(phi)
F/W = cosθ/sinφ
Unless I'm typing it wrong in excel
A2=5...
You mean just go directly with F/W = cos θ/sinφ and leave out this...
Plug in θ and get A
A^2=2^2+5^2-2(2)(5)cosθ
Then I'll plug in both θ and A to get Phi.
φ=arcsin[(5sin(θ)/A)]
If so, where would I get phi from if I'm only starting with θ?
I'm suppose to plug all of this into excel and make a plot with a line curving with θ being between 5 ≤ θ ≤ 60 degrees.
My idea was this:
Plug in θ and get A
A^2=2^2+5^2-2(2)(5)cosθ
Then I'll plug in both θ and A to get Phi.
φ=arcsin[(5sin(θ)/A)]
Then I'll plug θ and φ
F/W = cos θ/sinφ
But...
My plot didn't come out correctly, so I know I did it wrong. Where did I go wrong or was I even close?
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What is the true frequency of the uncalibrated generator?
The true frequency of the uncalibrated generator is 904 Hz.
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